Lifting of Characters for Nonlinear Simply

Laced Groups

Jeffrey Adams and Rebecca Herb

September 5, 2008

Abstract

One aspect of the Langlands program for linear groups is lifting of charac-ters, which relates virtual representations on a group G with those on anendoscopic group for G. The goal of this paper is to extend this theory tononlinear two-fold covers of real groups in the simply laced case. Suppose Gis a two-fold cover of a real reductive group G. The main result is that there

is an operation, denoted LifteGG, taking a stable virtual character of G to 0 or

a virtual genuine character of G, and LifteGG(Θπ) may be explicitly computed

if π is a stable sum of standard modules.

1 Introduction

The Langlands program is concerned with representation theory and auto-morphic forms of algebraic (linear) groups. Suppose G is a nonlinear group,for example the metaplectic group Mp(2n), the nontrivial twofold cover ofSp(2n). It is well known that such groups play an important role in auto-morphic forms, although they don’t fit into the formalism of the Langlands

2000 Mathematics Subject Classification: Primary 22E50, Secondary 05E99

The first author was supported in part by National Science Foundation Grant #DMS-

0554278.

1

program. The goal of this paper and of [3] is to extend some of the resultsof the Langlands program to certain nonlinear groups.

Let G be the real points of a connected reductive algebraic group. Char-acters of admissible representations of G have been studied extensively, andthe theory is reasonably complete. In particular if π is a discrete series rep-resentation the character Θπ of π is known explicitly [14]. The formula forΘπ implicitly comes from the theory of transfer and endoscopy, originatingin work of Shelstad and Langlands on the trace formula and automorphicforms. We would like to extend the theory of transfer of characters to non-linear groups.

Suppose G is a nonlinear two-fold cover of G, i.e. G can not be realized asa subgroup of GL(n,C). Examples include the metaplectic groupMp(2n,R),the unique connected two-fold cover of Sp(2n,R), and the twofold cover ofGL(n,R) of [18], [19]. There is substantial evidence that character theory

for genuine representations of G (those which do not factor to G) may bereduced to that of a linear group.

The case of GL(n) over any local field of characteristic zero has beenstudied by Kazhdan, Patterson and Flicker ([9], [8], [20]). The general phi-losophy is spelled out in [21]. Some discussion of how to extend these ideasto general groups is given in [5], and this is carried out for G = Sp(2n,R) in[4].

Here is an outline of the approach. For now suppose G is the F-points of aconnected, reductive algebraic group defined over a local field of characteristic0, and G is a nonlinear two-fold cover of G. Identify the kernel of the coveringmap p : G → G with ±1. We would like to relate genuine characters of Gto characters of G. By character we mean the character of a representation,viewed as a function on the regular semisimple elements.

The theory of endoscopy for G relates characters of G to characters ofendoscopic groups of smaller semisimple rank. A key part of the theory is arelationship between semisimple conjugacy classes in G and H .

Our theory is modelled on this, with G in place of G, and G playingthe role of H . We define a relation on conjugacy classes as follows. Forg ∈ G define ψ(g) = s(g)2 where s(g) ∈ G satisfies p(g) = g. Then ψ(g)is independent of the choice of s(g), and this induces a map from conjugacy

classes of G to those of G. This idea goes back to [9].Now suppose π is an admissible representation of G, with character Θπ

viewed as a function on the (strongly) regular semisimple elements G′ of G

2

(see Section 3). For g ∈ G′ define

(1.1) ψ∗(Θπ)(g) =∑

{h∈G |ψ(h)=eg}

Θπ(h).

This is a conjugation invariant function on G′. One can ask whether it is thecharacter of a genuine representation π of G. More generally it may be a vir-tual character, i.e. the character of a finite sum of irreducible representationswith (possibly negative) integral coefficients.

Although this definition is too naive for several reasons, we describe abasic property of characters which suggest it is at least a good first approx-imation. Suppose H is a Cartan subgroup of G, with inverse image H inG. Then H is not necessarily abelian, and we let Z(H) be its center. The

fact that H is not abelian is a major reason why the representation theoryof G at least appears to be more complicated than that of G. However, atleast as far as character theory goes, the situation is actually very simple: thecharacter Θeπ of an admissible genuine representation of G satisfies Θeπ(g) = 0

for g 6∈ Z(H) (Lemma 9.3). It is easy to see that ψ(H) ⊂ Z(H), of finiteindex, so at least from this point of view (1.1) is similar to the character of

a genuine representation of G.An obvious shortcoming of(1.1) is that it is not necessarily the case that

ψ∗(Θπ)(−g) = −ψ∗(Θπ)(g), an obvious requirement if ψ∗(Θπ) is to be thecharacter of a genuine representation. It would be better to define φ(h) = h2,sum over h satisfying φ(h) = p(g), and modify the definition by some genuine

function µ of G:

(1.2) φ∗(Θπ)(g) =∑

{h∈G |φ(h)=p(eg)}

µ(g)Θπ(h).

We then need to choose µ appropriately.There are also some Weyl denominators to take into account. Suppose

Φ+ is a set of positive roots of H in G. For h ∈ H one version of the Weyldenominator is |D(h)| 12 = |∏α∈Φ+(1 − α−1(h))||eρ(h)| (cf. (5.1)). This isindependent of Φ+ and is well defined. The character of any representationhas a particular form when muliplied by |D(h)| 12 ; for the left hand side of(1.1) to have a chance of having this form we should multiply each term on

the right by the quotient of Weyl denominators |D(h)| 12/|D(g)| 12 .

3

Putting these two considerations together we look for a function ∆ onG′ × G′ satisfying the following conditions:

(1.3)

∆(h, g) = 0 unless p(g) = φ(h)

|∆(h, g)| = |D(h)| 12/|D(g)| 12∆(xhx−1, xgx−1) = ∆(h, g) (x ∈ G, x = p(g))

∆(h,−g) = −∆(h, g).

Given such a function we define

(1.4) LifteGG(Θπ)(g) =

∑

{h∈G |φ(h)=p(eg)}

∆(h, g)Θπ(h).

Now LifteGG(Θπ) is a conjugation invariant, genuine function on G, and is a

reasonable candidate for the character of a virtual representation. The lattercondition amounts to further conditions on ∆. If this holds, one can define

LifteGG(π) to be the virtual representation of G whose character is equal to

LifteGG(Θπ): Θ

LifteGG(π)

= LifteGG(Θπ).

For a version of this for covering groups of GL(n) see [20].Formula (1.3) is analogous with the main character identity in the theory

of endoscopy. For example see [25]. In that setting ∆ is a transfer factor, andby analogy we use the same terminology here. As in the theory of endoscopy,correctly defining transfer factors is a difficult part of the theory.

We now discuss another less obvious consideration which arises. RecallΘπ is conjugation invariant: Θπ(g) = Θπ(g

′) if g, g′ ∈ G′ and g′ = xgx−1 forsome x ∈ G. Following Langlands and Shelstad we say π and Θπ are stableif the following stronger condition holds: Θπ(g) = Θπ(g

′) if g, g′ ∈ G′ andg′ = xgx−1 for some x ∈ G(F).

For GL(n) every conjugation invariant function is automatically stable,so stability plays no role in [8], [9] and [20].

For guidance we consider the case of G = Sp(2n,R), G = Mp(2n,R),as discussed in [4]. Let G′ = SO(n + 1, n). In this case the map φ aboveis replaced by a bijection between (strongly) regular semisimple conjugacyclasses in G and G′. Suppose π is a stable representation of G′. The mainresult of [4] is that, for appropriate definition of the transfer factor ∆, if wedefine

(1.5) LifteGG′(Θπ)(g) = ∆(h, g)Θπ(h) (φ(h) = p(g))

4

then LifteGG′(Θπ) is the character of a genuine virtual representation of G.

This suggests we should relate characters of G with stable characters ofreal forms of the dual group G∨(C). We expect the general theory to have thisform. In this paper we restrict ourselves to simply laced groups. This avoidsa number of technical complications, and the distinction between G(C) andG∨(C) is less critical. We may now state a special case of the main result.

Theorem 1.6 (Theorem 19.1 and Corollary 19.11) Suppose G(C) is aconnected, reductive, simply laced complex group, with real points G. Weassume the derived group of G(C) is acceptable. Suppose G is a admissiblecover of G (cf. Section 2 and Definition 3.3). Then we can define the transferfactor∆(h, g), satisfying (1.3), such that for all stable admissible representa-tions π of G,

(1.7) LifteGG(Θπ)(g) =

∑

{h∈G |φ(h)=p(eg)}

∆(h, g)Θπ(h)

is the character of a genuine virtual representation π of G, or 0. We say π

is the lift of π, and write π = LifteGG(π).

If π is a stable sum of standard modules we compute LifteGG(π) explicitly.

We note that unlike the case of Mp(2n,R), the role of stability is verysubtle in the simply laced case. It appears in the proof of Theorem 14.1; seeRemark 14.32.

Theorem 1.6 is formally similar to transfer in the setting of endoscopic

groups. For example see [25, Lemma 4.2.4]. More precisely LifteGG is analogous

to the simplest case of endoscopy: transfer from the quasisplit form Gqs of GtoG [24]. As we will see below it is often possible to obtain a single irreducible

representation of G as a lift, and there is no natural notion of stability forG. This suggests that in the simply laced case this is the only notion oflifting which is needed. We note that there is a useful notion of stabilityfor Mp(2n,R) [4], and David Renard has defined a family of “endoscopicgroups” for Mp(2n,R) and proved lifting results for them [23]. We believethis is the only situation in which this is either necessary or possible. Weplan to return to the two root length case in another paper.

Example 1.8 Let G = SL(2,R) and let G be the unique non-trivial twofolder cover of G. Let B be a Borel subgroup of G and write B = AN with

5

A ≃ R×. For δ = ±1 and ν ∈ C define a character χ of R× by χ(−1) = δand χ(x) = xν for x ∈ R+. Let π(δ, ν) be the corresponding principal seriesrepresentation of G, i.e. IndGAN(χ⊗ 1) (normalized induction).

Now let A be the inverse image of A in G. Then A ≃ R× ∪ iR×. We mayidentify the inverse image of N with N , and let B = AN . For ǫ ∈ ±1 andγ ∈ C define χ(i) = ǫi and χ(x) = xγ for x ∈ R+. Let π(ǫ, γ) be the genuine

principal series representation IndeGeB(χ⊗ 1) of G.

Let π = π(δ, ν). It is easy to see that LifteGG(Θπ) = 0 if δ = −1. If

δ = 1 then by an easy calculation (see Section 11), using the transfer factorof Section 5 we see

(1.9) LifteGG(π(1, ν)) = π(1, ν/2) ⊕ π(−1, ν/2).

Note that the image of φ restricted to A is A0, and the character of this Liftis 0 on p−1(−A0) = iR×.

This example illustrates several features. In this example computing

LifteGG(π) essentially reduces to computing Lift

eAA(χ). This is a special case

of the general situation.Note that π(±1, ν/2) have different central characters (the center of G

is Z(G) ≃ Z/4Z), so LifteGG(π(1, ν)) does not have a central character. Also

note that this character is supported on A0, which is a proper subgroup ofZ(A).

A hint that Theorem 1.6 is not the best result possible is seen by consid-ering the preceding example from another point of view. View SL(2,R) asSp(2,R), and apply [4]. In this case G′ = SO(2, 1). There are two principalseries representations π(±1, ν) of SO(2, 1) analogous to those described forSL(2,R). Writing Lift for the lifting of [4] we see

(1.10) LiftMp(2,R)SO(2,1) (π(±1, ν)) = π(±1, ν).

See [4, Proposition 15.10]. Thus we may obtain each principal series repre-

sentation π(±1, ν) of SL(2,R), rather than just their sum as in (1.9).Note that SO(2, 1) is isomorphic to PSL(2,R), the real points of PSL(2,C)

(also denoted PGL(2,R)). This is the real form of the adjoint group, andsuggests it should be possible to generalize lifting of (1.5) to allow G to be

replaced with a real form of a quotient of G(C) (without changing G). Werevisit the previous example from this point of view.

6

Example 1.11 Let G(C) = SL(2,C) and

G(C) = PSL(2,C) = SL(2,C)/± I.

Let G = SL(2,R) and let G = PSL(2,R) be the real points of G(C). RecallG ≃ PGL(2,R) ≃ SO(2, 1).

We define an orbit correspondence between G and G as follows. Forg ∈ G choose an inverse image s(g) of g in G(C), and let φ(g) = s(g)2. Thisis independent of the choice of s(g), and φ(g) ∈ SL(2,R).

Let A(C) = {diag(z, 1/z)} ⊂ G(C), and let A(C) be its image in G(C).Then A(C) and A(C) are defined over R, and let A = A(R), A = A(R). BothA and A are isomorphic to R×. Note that while the map A(C) → A(C) issurjective, the restriction A(R) → A(R) is not.

Write diag(z, 1z) for the image of diag(z, 1

z) in A(C). Let g = diag(x, 1

x) ⊂

A with x ∈ R×. Then φ(g) = diag(x2, 1x2 ) ∈ A0. However suppose y ∈ R×

and let g = diag(iy, 1iy

) ∈ A. Then φ(g) = diag(−y2,− 1y2

). Therefore (unlike

in Example 1.8) φ maps A onto A.

This suggests that if we develop a similar lifting theory from G to G, thenthe lift of a principal series representation will have support on all of A. In

fact this is the case: we can define LifteGG. We recover (1.10) from this point

of view (the difference between ν and ν/2 on the right hand side is an issueof normalization):

(1.12) LifteGG(π(±1, ν)) = π(±1, ν/2).

Motivated by this example, we generalize (1.5) as follows. Let G(C) be

our given complex group, with real points G and nonlinear cover G. SupposeC ⊂ Z(G) is a two group and let G(C) = G(C)/C, with real points G. Forg ∈ G define φ(g) = s(g)2 where s(g) is an inverse image of g in G(C).The fact that C is a two-group implies φ(g) ∈ G, and this is independentof the choice of s(g). It is not necessarily the case that φ induces a map onconjugacy classes; however it does define a map on stable conjugacy classes,which is consistent with our application.

We need to make several technical assumptions on C (Definition 3.14);C = 1 is allowed. Under these assumptions we can define the transfer factor

∆(h, g) on G′ × G and define the lift Lift

eGG(π) of a stable representation of G

as in (1.5). See Definition 9.10.

7

Now suppose H is a Cartan subgroup of G, with inverse image H in Gand corresponding subgroup H ⊂ G. Our assumptions on C imply φ(H) ⊂p(Z(H)), which is necessary to have a meaningful theory. As in the preceding

examples, if this image is large then LifteGG(π) will have fewer terms in its sum.

This is desirable; it would be nice to have only one term in the sum. TakingC larger makes this image larger, so we would like to take C as large aspossible. An optimal choice would be to choose C so that φ(H) = p(Z(H))for all Cartan subgroups, in which case each Lift would consist of a singleterm. This is not possible in general. See Example 3.16.

The proofs follow the same general outline as those of [4]. A hard partof the theory is the definition of transfer factors. Once transfer factors andlifting are defined we first show that the lift of a stable invariant eigendistri-bution is an invariant eigendistribution. This requires checking that liftingrespects the Hirai matching conditions, and it is here that stability plays acrucial role. Once this is done it is fairly easy to compute the lift of a stablesum of discrete series representations. It is also straightforward to prove thatlifting commutes with parabolic induction. This enables us to compute thelift of any stable sum of standard modules. In principal, giving the Kazhdan-Lusztig-Vogan polynomials for G and G, we may then compute the Lift ofany stable virtual representation of G.

Here is an outline of the contents of the paper.In Section 2 we make some basic definitions and establish some notation.

Admissible triples (G, G,G) are defined in Section 3. Such a set consists of

a nonlinear group G, a linear group G, and the real form G of a quotientof G(C), satisfying certain assumptions. Section 3 also contains a discussionof some basic structural facts about Cartan subgroups, and defines the or-bit correspondence φ. Section 4 recalls some standard facts about Cartansubgroups and Cayley transforms, and generalizes them to nonlinear groups.Cayley transforms are an important tool in the theory.

The basic, and most important case, that of the real points of a semisim-ple, simply connected complex group, is discussed in Section 5. The transferfactors are canonical in this case, have a simple form, and this case providesguidance for the general theory. We recommmend the reader restrict to thiscase the first time through.

Sections 6 and 7 are technically the most difficult. Section 6 defines cer-tain characters of Cartan subgroups and their application to transfer factors.

8

Transfer factors are defined in Section 7. Some constants associated to Car-tan subgroups are defined and studied in Section 8.

With transfer factors in place we define lifting and study its basic prop-erties in Section 9. The case of tori, which is both a good example and animportant special case, is discussed in Section 10. Minimal principal seriesof split groups are covered in Section 11, and discrete series on the compactCartan in Section 12. Section 13 summarizes some material about invarianteigendistributions which we will need. Some details of an extension of Hirai’sresults which we need are in the Appendix (Section 20); this is work of thesecond author.

In Section 14 we use the results of Section 13 to prove that the lift of astable invariant eigendistribution in an invariant eigendistribution.

We prove that lifting commutes with parabolic induction in Section 15.Modified character data appropriate to our setting is defined in Section 16.This is a (mild) modification of character data and the Langlands classifica-tion due to Vogan [26]. Formal lifting of (modified) character data is definedin Section 17. This essentially comes down to lifting applied to a Cartansubgroup.

We compute the lifting of stable discrete series representations in Section18, and of general standard modules in Section 19.

These results are closely related to duality of representations, also knownas Vogan duality, introduced in [27]. Duality for nonlinear covers of simplylaced groups is discussed in a paper by Peter Trapa and the first author [3].These two results grew up together, and we thank Peter Trapa for manyhelpful discussions.

2 Some Notation

A simple root system is said to be simply laced if all roots have the samelength, and an arbitrary root system is simply laced if this holds for eachsimple factor. More succinctly a root system is simply laced if whenever α, βare non-proportional roots then 〈α, β∨〉 = 0,±1. We adopt the conventionthat in this case all roots are long.

We say a root system is oddly laced if whenever α, β are non-proportionalroots then 〈α, β∨〉 = 0 or is odd. Thus oddly laced is shorthand for each simplefactor is simply laced or of type G2. We also adopt the convention that intype G2 all roots are long. The reason for these conventions is Lemma 3.2.

9

The main results in this paper hold for oddly laced groups, although (withthe general case in mind) we will only make this assumption when necessary.

Suppose G(C) is a connected, reductive complex Lie group. Let Gad(C)be the adjoint group. If G(C) is defined over R let G be its real points, andlet Gad be the real points of Gad(C). Write int(g) for the action of Gad(C)on G(C), or Gad on G.

Let Gd(C) be the derived group of G(C), and let Gd be the derived groupof G. Note that by (2.8)(b) Gd is the identity component of the real pointsof Gd(C).

We denote real Lie algebras by Gothic letters h, g, t, . . . , and their com-plexifications by h(C), g(C), t(C), . . . . We write σ for the action of the non-trivial element of the Galois group on g(C) and G(C), so g = g(C)σ andG(R) = G(C)σ.

Fix a Cartan involution θ of G(C), i.e. K = Gθ is a maximal compactsubgroup of G. Unless otherwise noted all Cartan subgroups of G or G(C)are assumed to be θ-stable. Let H be a (θ-stable) Cartan subgroup of G,with complexification H(C). Write h = t ⊕ a as usual, and H = TA withT = H ∩K and A = exp(a). Let Φ = Φ(G,H) be the root system of H(C)in G(C). Let W (G,H) = NormG(H)/H ; this is the real Weyl group. Itis a subgroup of the absolute Weyl group W = W (G(C), H(C)) which isisomorphic to the Weyl group of Φ. The Cartan involution θ acts on W , andW (G,H) ⊂W θ.

Note σ(α) = −θ(α) for all α ∈ Φ. Roots are classified as real, imaginary,complex, or compact as in [26]. Write Φ = Φr ∪ Φi ∪ Φcx accordingly; notethat Φr and Φi are root systems. We also have the decomposition of Φi =Φi,c ∪ Φi,n into compact and noncompact roots; Φi,c is a root system. If Φ+

is a set of positive roots, write Φ+r = Φ+ ∩ Φr, and Φ+

i ,Φ+cx similarly. Let

ρ = ρ(Φ+) = 12

∑α∈Φ+ α as usual, and define ρr, ρi and ρcx similarly, so

ρ = ρr + ρi + ρcx.We say G(C) is acceptable if ρ exponentiates to a character of H(C).An important role is played by certain sets of positive roots:

Definition 2.1 A set Φ+ of positive roots is said to be special if σ(Φ+cx) =

Φ+cx, or equivalently σ(α) > 0 for all positive non-imaginary roots.

Let

(2.2) Γ(H) = exp(ia) ∩H = {exp(iX) |X ∈ a, exp(2iX) = 1}.

10

An important role is played by a certain character of Γ(H). Let S be a setof complex roots such that the set of all complex roots is {±α,±σα |α ∈ S}.Define

(2.3) ζcx(G,H)(h) =∏

S

eα(h) (h ∈ Γ(H)).

If G,H are understood we let ζcx = ζcx(G,H). It is elementary to see thatζcx is independent of the choice of S, factors to Gad, and for all h ∈ Γ(H)satisfies

ζcx(h) = ζcx(wh) (w ∈W (G,H))(2.4)(a)

ζcx(h) = eρcx(h) = eρ−ρr(h)(2.4)(b)

for any special set of positive roots. For (b) note that if Φ+ is a special setof positive roots and S ⊂ Φ+

cx is as above, then for X ∈ a(C)

(2.5) ρcx(X) = (ρ− ρr)(X) =∑

α∈S

α(X),

and it follows that eρcx(exp(X)) = eρ−ρr(exp(X)) = eρcx(X) is a well-definedcharacter of A(C).

For α ∈ Φr let mα = α∨(−1) = exp(πiα∨) and define

(2.6) Γr(H) = 〈mα |α ∈ Φr〉 ⊂ Γ(H) ∩G0d.

It is well known that

(2.7) H = Γ(H)H0, H ∩G0 = Γr(H)H0.

It is also well known that if Hs is a maximally split Cartan subgroup of Gthen

(2.8)(a) G = HsG0,

and this implies

(2.8)(b) Gd = G0d.

11

3 Admissible Triples

Fix G as in Section 2 and suppose p : G→ G is a two-fold cover. We identifythe kernel of p with ±1.

If H is a subgroup of G we let H = p−1(H). Let Z(H) be the center of H

and let Z0(H) = p(Z(H)). It is immediate that Z(H) = p−1(Z0(H)), so to

describe Z(H) it is enough to describe Z0(H). In particular Z0(G) ⊂ Z(G)plays an important role; it is immediate that Z0(G) = Z(G) if G is connected(cf. (4.4)).

Suppose H is a Cartan subgroup of G. Typically H is not abelian andZ0(H) plays an important role. It is easy to see

(3.1) H0 ⊂ Z0(H) ⊂ H.

Fix a real or imaginary root α. Associated to α is the root subgroup Mα,which is locally isomorphic to SL(2,R) or SU(2). As in [3, Definition 3.2]we say α is metaplectic if p−1(Mα) is a nonlinear group. If Mα is compactit has no such cover, nor does SL(2,R)/ ± 1, so if α is metaplectic thenMα ≃ SL(2,R) and α is either real or noncompact imaginary.

Let mα be an inverse image of mα in G. It is easy to see that mα hasorder 4 if α is metaplectic, and 1 or 2 otherwise. For the next Lemma see [6]or [3, Lemma 3.3].

Lemma 3.2 Assume G(C) is simple and simply connected and that G isnonlinear. Fix a Cartan subgroup H of G. Then a real or imaginary root α ofH(C) in G(C) is metaplectic if and only if α is long and is real or noncompactimaginary. Furthermore G admits a nonlinear cover if and only if there is aCartan subgroup H with a long real or long noncompact imaginary root. Ifthis condition holds the nonlinear two-fold cover is unique up to isomorphism.

It is enough to check this condition on a fundamental or maximally splitCartan subgroup. If G is oddly laced it is enough to check this condition onany Cartan subgroup.

Definition 3.3 We say that p : G → G is an admissible two-fold cover iffor every θ-stable Cartan subgroup H, every long real or long noncompactimaginary root is metaplectic.

Equivalently G is admissible if and only if Gi is nonlinear for every simplefactor Gi of G which admits such a cover.

12

See [3, Definition 3.4]).

Example 3.4 Let G = U(1, 1). This has three non-trivial two-fold covers,described by their restriction to T ≃ U(1), the diagonal maximal compact

subgroup. The cover of T is a connected torus T , and is best described byits character lattice. Write the character lattice X∗(T ) as Z2 in the usual

coordinates. The three covers are given by X∗(T ) = 12Z ⊕ Z, Z ⊕ 1

2Z, or

Z2 ∪ (Z + 12)2.

The derived group is SU(1, 1) ≃ SL(2,R), which has a unique non-trivial two-fold cover. The first two covers of U(1, 1) restrict non-triviallyto SU(1, 1), and are therefore admissible. The third one is trivial whenrestricted to SU(1, 1) (it is the

√det cover), and is not admissible.

As discussed in the Introduction we are going to lift characters from areal form of a quotient of G(C). We need to impose some conditions on thisquotient. This will take up the remainder of this section.

Suppose C is a finite subgroup of Z(G). Then G(C) = G(C)/C is definedover R, and letG be its real points. Write p : G(C) → G(C) for the projectionmap. Let Orb(G) be the conjugacy classes of G. If g, g′ ∈ G are conjugateby G(C) we say g, g′ are stably conjugate, and let Orbst(G) be the set ofstable conjugacy classes. (This is a naive definition, which agrees with theusual one for strongly regular semisimple elements.) Similar notation appliesto other groups.

For g ∈ G let O(G, g) be the conjugacy class of g, and Ost(G, g) ={xgx−1 | x ∈ G(C), xgx−1 ∈ G} the stable conjugacy class. Similar notationapplies to other groups.

Definition 3.5 Assume C is a two-group. For h ∈ G let φ(h) = s(h)2 wheres : G(C) → G(C) is any section.

Lemma 3.6 The map φ is well defined, and satisfies:

1. φ(G) ⊂ G and φ(G0) ⊂ G0,

2. φ induces a map Orb(G0) → Orb(G0),

3. φ induces a map Orbst(G) → Orbst(G).

13

Proof. Since C is a two-group it is immediate that φ(h) is independent ofthe choice of s.

Suppose h ∈ G, g ∈ G(C), and p(g) = h. Then p(σ(g)) = σ(p(g)) = p(g),so σ(g) = zg for some z ∈ C. Since C is a two-group σ(g)2 = g2, so

φ(h) = g2 ∈ G. Furthermore since p : G0 → G0

is surjective the secondassertion in (1) is clear.

Define

(3.7) φ(O(G0, g)) = O(G0, φ(g)) (g ∈ G

0).

If x ∈ G0

choose y ∈ G0 with p(y) = x. Then φ(xgx−1) = yφ(g)y−1 so thisis well defined, proving (2). Similarly define

(3.8) φ(Ost(G, g)) = Ost(G, φ(g)) (g ∈ G).

Suppose x ∈ G(C) and xgx−1 ∈ G. Choose y ∈ G(C) with p(y) = x. Thenφ(xgx−1) = yφ(g)y−1, so again φ is well-defined. �

It is worth noting that φ does not define a map Orb(G) → Orb(G).Suppose we try to define φ(O(G, g)) = O(G, φ(g)) as in (3.7). For this tobe well defined we need to know that for x ∈ G, φ(xgx−1) is G-conjugateto φ(g). This is true if x ∈ p(G), but not in general. (By (3) φ(g) is stablyconjugate to φ(xgx−1)).

Example 3.9 Let G = SL(2,R) and G = PSL(2,R) ≃ SO(2, 1) (cf. Ex-ample 1.11). Let t(θ) = diag(t′(θ), 1) ∈ SO(2, 1) (with the appropriate form)

where t′(θ) =

(cos(θ) sin(θ)− sin(θ) cos(θ)

)∈ SL(2,R). Then just as in Example 1.11

φ(t(θ)) = t′(θ). However while t(θ) is conjugate to t(−θ), t′(θ) is not con-jugate to t′(−θ) (for generic θ). Therefore the map φ(O(SO(2, 1), t(θ)) =O(SL(2,R), t′(θ)) is not well defined.

We assume throughout this paper that C is a two-group (and thereforefinite).

Now fix a Cartan subgroup H of G, and let H(C) = H(C)/C ⊂ G(C),with real points H . We define φ : H(C) → H(C) by the same formula as inDefinition 3.5. Equivalently, if we write exp : h(C) → H(C), exp : h(C) →H(C), then for X ∈ h(C), φ(exp(X)) = exp(2X). It is immediate that

(3.10) α(φ(h)) = α(h)2 (for all α ∈ Φ).

14

Lemma 3.11 The homomorphism φ : H(C) → H(C) has the followingproperties.

1. φ(wh) = wφ(h) for all w ∈W (Φ), h ∈ H(C),

2. φ(H0) = H0 ,

3. φ(h) = 1 for all h ∈ Γr(H),

4. φ(H ∩G0) = H0 ,

5. φ(Γ(H)) = Γ(H) ∩ C,

6. φ(H) = (Γ(H) ∩ C)H0.

Proof. The first two are clear from the definition. For (3) note thatφ(exp(πiα∨)) = exp(2πiα∨) = 1 for all α ∈ Φr. Then (4) follows since

H ∩G0= Γr(H)H

0(cf. 2.7).

Let h = exp(iX) ∈ Γ(H) where X ∈ a with exp(2iX) = 1. Now φ(h) =exp(2iX) ∈ exp(ia). But p(φ(h)) = exp(2iX) = 1 so that φ(h) ∈ C ∩exp(ia) ⊂ Γ(H)∩C. Conversely, let zG = exp(iY ) ∈ C ∩ Γ(H) where Y ∈ a

with exp(2iY ) = 1. Since zG ∈ C, p(zG) = exp(iY ) = 1. Let z = exp(iY/2).

Then z ∈ Γ(H) and φ(z) = zG. Finally (6) follows since H = Γ(H)H0. �

It is a standard fact that the character of an irreducible genuine repre-sentation of G, considered as a function on the regular semisimple elements,vanishes off of Z(H) (Lemma 9.3). Our character identities involve the imageof φ in H , so we would like to know that φ(H) ⊂ Z0(H), with image as largeas possible.

Recall C ⊂ Z(G). We first show that we may as well assume C ⊂ Z0(G).Let Hs be a maximally split Cartan subgroup of G, and assume φ(Hs) ⊂

Z0(Hs). Let Cs = Γ(Hs) ∩ C so that by Lemma 3.11 CsH0s = φ(Hs) ⊂

Z0(Hs). Therefore Cs centralizes Hs, and by (4.3) Cs centralizes G0. There-

fore Cs centralizes HsG0 which equals G by (2.8)(a). Therefore Cs ⊂ Z0(G).Now for a general Cartan subgroup H = TA, there is a maximally split

Cartan Hs = TsAs so that A ⊂ As. Then Γ(H) ⊂ Γ(Hs), so that φ(H) =(C ∩Γ(H))H0 = (Cs ∩Γ(H))H0. That is, if we use Cs in place of C we havethe same images φ(H) for every Cartan subgroup H . Thus we may as wellassume that C ⊂ Z0(G).

15

Lemma 3.12 Assume that C ⊂ Z0(G). Then for every Cartan subgroupφ(Γ(H)) ⊂ Z0(G) and φ(H) ⊂ Z0(H). Furthermore φ(Z(G)) ⊂ Z0(G).

Proof. The first part follows from the above discussion. Let Hs be a max-imally split Cartan subgroup with roots Φ. Let z ∈ Z(G) ⊂ Hs, z ∈ Z(Hs)such that p(z) = φ(z). Then for every α ∈ Φ,

α(z) = α(φ(z)) = α(z)2 = 1

by (3.10). Thus z ∈ Cent eG(G0) ∩ Z(Hs) = Z(G) as above. �

To reiterate, so far we have assumed C is a (finite) subgroup of Z0(G),and z2 = 1 for all z ∈ C. For the definition of transfer factors we need toimpose a further technical condition on C.

Suppose χ is a genuine character of Z(G). Then χ2 factors to a characterof Z0(G). Suppose z ∈ Z0(G) has order 2 and p(z) = z. Then z has order 2or 4 and χ2(z) = 1 or −1 accordingly, independent of χ. Assume that Gd(C)is acceptable.

Let

(3.13) ζ2(z) = χ2(z)eρ(z) z ∈ Z0(G) ∩Gd, z2 = 1.

This is well defined since Gd is acceptable, and is independent of thechoices of χ, H and Φ+.

Definition 3.14 An admissible triple is a set (G, G,G) where:

1. G is the set of real points of a connected reductive complex Lie groupG(C) such that the derived group Gd(C) is acceptable, with oddly lacedroot system i.e. each simple factor is simply laced or G2 (cf. Section2).

2. p : G→ G is an admissible two-fold cover (Definition 3.3).

3. G is the set of real points of G(C) = G(C)/C where C is a finitesubgroup of Z(G) satisfying the following conditions:

(a) c2 = 1 for all c ∈ C,

(b) C ⊂ Z0(G) = p(Z(G)),

(c) ζ2(z) = 1 for all z ∈ C ∩Gd (cf. 3.13).

16

For example if G and G satisfy (1) and (2) then (G, G,G) is an admissibletriple.

More generally, suppose G and G satisfy conditions (1) and (2). Then

(G, G,G) is an admissible triple for G(C) = G(C)/C for any subgroup Csatisfying conditions (3)(a-c). We may take C = 1; by Lemma 3.11(6) takingC bigger makes φ(H) ⊂ Z0(H) bigger, and in general, we get the best liftingresults when we take C as large as possible. However, even if we take C aslarge as possible for G, it may not be the maximal choice for Levi subgroupsof G. Thus we do not specify C beyond the requirements of Definition 3.14.

Assume that (G, G,G) is an admissible triple. Let H = TA be a Cartansubgroup of G. Let M = CentG(A) be a cuspidal Levi subgroup. Then

M ⊂ G and M = CentG(A) ⊂ G are also cuspidal Levi subgroups. We will

show in Section 15 that (M,M,M) is also an admissible triple. Thus ourconstructions are compatible with parabolic induction.

The following lemma, which is an immediate consequence of (6.21) isuseful for producing admissible triples.

Lemma 3.15 Let H be a maximally split Cartan subgroup of G. Thenζ2(c) = 1 for all c ∈ Z0(G) ∩ Γr(H).

Lemma 3.16 Suppose that G(C) is simple and simply connected, and let Hbe a maximally split Cartan subgroup of G. Then we can choose C so thatφ(H) = Z0(H) except when G = SU(n, n) where n is even, G = Spin(p, q)where p and q are even with p > q ≥ 2 or when G = Spin∗(2n) where n ≡ 0mod 4.

Proof.Clearly if Z0(H) is connected we can take C = {1}. We will prove in (4.4)

and Proposition 4.7 that Z0(H) = Z0(G)H0 = Z(G)H0 since G is connected.But since H = Γr(H)H0 where Γr(H) is a two group, it is easy to see thatZ(G)H0 = Ze(G)H0 where Ze(G) is the group of elements in Z(G) with evenorder. Thus Z0(H) is connected when Ze(G) = {1}. This will be the casefor all real forms when Φ is of type A2n, E6, E8, or G2. Thus we may as wellassume that Φ is type A2n−1, Dn, or E7. In these cases Ze(G) = Z(G). Wemay also assume that G is a real form such that Z0(H) = Z(G)H0 is notconnected. In particular, we can assume that G is not compact or complex.Suppose that Z(G) ⊂ Γr(H). Then Z(G) is a two-group, and by Lemma3.15 we can take C = Z(G), giving us φ(H) = Z(G)H0 = Z0(H) as desired.This will always be the case if G is the split real form.

17

Suppose that Φ is of type A2n−1. Then G is split or H is connectedunless G = SU(n, n). In this case Z0(H) = H has two components. NowZ(G) =< z > is cyclic of order 2n and z0 = zn ∈ Γr(H) is the unique elementof order two. Thus the biggest C we can take is C = {1, z0}. When n is odd,z0 6∈ H0 so that φ(H) = CH0 = H , but when n is even, z0 ∈ H0 so thatφ(H) = CH0 = H0 is a proper subgroup of Z0(H).

Suppose that Φ is of type Dn. Let G = Spin(p, q) with p ≥ q. Then G issplit or H is connected unless p > q ≥ 2. If p, q are odd, then Z(G) = Z2 ⊂Γr(H). If p, q are even, then Z(G) = {1, z0, z1, z0z1} where z0 ∈ Γr(H), butz1 6∈ Γ(H). Thus the biggest C we can take is C = {1, z0}. But z1 6∈ CH0

so that CH0 is a proper subgroup of Z0(H).Let G = Spin∗(2n). If n is odd H is connected, so assume n = 2k. Then

Z(G) = {1, z0, z1, z0z1} as above where z1 ∈ Γr(H), but z0 6∈ Γ(H) so thebiggest C we can take is C = {1, z1}. If k is even, then z0 6∈ CH0 whilez0 ∈ CH0 when k is odd.

Finally, if Φ = E7, then either G is split or Z0(H) is connected. �

Example 3.17 Let H = TA be a maximally split Cartan subgroup of G.By Lemma 3.16 it is not always possible to pick C satisfying the conditionsof Definition 3.14 such that φ(H) = Z0(H). Let M = CentG(A). Here weshow it is possible to pick a finite central subgroup CM of M satisfying theconditions of Definition 3.14 for M such that φ(H) = Z0(H) where φ isdefined using M(C) = M(C)/CM .

Recall H = Γ(H)H0, so H = Γ(H)H0 and Z(H) = Z(Γ(H))H0. ThenZ0(H) = Z0(Γ(H))H0, and since Γ(H) is a two-group there exists CM ⊂Z0(Γ(H)) such that CM ∩H0 = {1} and Z0(H) = CMH

0. We have to showCM satisfies the conditions of Definition 3.14(3). Condition (a) is obvious,and sinceM = HM0 = Γ(H)H0M0 it follows easily that Z0(Γ(H)) ⊂ Z0(M),which is (b). Finally CM ∩Md ⊂ CM ∩H0 = {1}, giving (c).

4 Cartan Subgroups and Cayley Transforms

We need some structural facts about Cartan subgroups. We first definecommutators on G with respect to an admissible cover G.

Assume for the moment that Gd(C) is simply connected and τ is anautomorphism of G. Then τ stabilizes Gd = G0

d (cf. (2.8)(b)) and by Lemma3.2 it lifts uniquely to an automorphism τ of the (unique) admissible cover

18

Gd of Gd. Suppose g ∈ Gd and τ(g) = g. Choose an inverse image g of g andset

(4.1) {τ, g} = τ (g)g−1.

Then p({τ, g}) = 1 so {τ, g} = ±1. It is independent of the choice of g.We now drop the assumption thatGd(C) is simply connected, but suppose

that G is an admissible cover of G. Suppose g, h are commuting elements ofG. Choose inverse images g, h of g, h in G and define

(4.2) {g, h} = ghg−1h−1.

Since p(ghg−1h−1) = ghg−1h−1 = 1, {g, h} = ±1, independent of the choices.

We say {g, h} is the commutator of g, h (with respect to the cover G). It iscontinuous in each factor so

(4.3) {Z(G), G0} = 1.

Note that this implies

(4.4) Z(G) = Z(G) if G is connected.

In general we only have Z(G) ⊂ Z(G), for example if G is a torus and G isnonabelian.

Suppose α ∈ Φ(G,H) is a long real root. Let Mα be as in Section 3.By Lemma 3.2 α is metaplectic, and by the discussion at the beginning ofSection 3, Mα ≃ SL(2,R). Let τ be an automorphism of Gd satisfyingτ(H ∩ Gd) = H ∩ Gd and τ(α) = α. Then τ(Mα) = Mα, so for g ∈ Mα

define {τ, g} by (4.1) applied to Mα . Let Tα ≃ S1 be a τ -stable compactCartan subgroup of Mα. Then for all t ∈ Tα, τ(t) = t or t−1. Recallmα = α∨(−1) ∈ Z(Mα).

For the next result we do not need to assume G is oddly laced.

Proposition 4.5 Suppose α is a long real root. Then

(4.6)(a) {τ,mα} =

{1 τ(t) = t for all t ∈ Tα

−1 τ(t) = t−1 for all t ∈ Tα

19

For all h ∈ H,

(4.6)(b) {h,mα} = sgn(α(h)).

If β ∈ Φr(G,H) then

(4.6)(c) {mα, mβ} = (−1)〈α,β∨〉.

Proof. We can compute τ(mα) by working in Tα. If τ acts trivially on Tαthen the same holds for the action of τ on Tα and {τ,mα} = 1. Suppose

τ(t) = t−1 for all t ∈ Tα. Then τ(t) = t−1 for all t ∈ Tα. Let mα be an

inverse image of mα in Tα. Then {τ,mα} = m−2α . Since α is metaplectic mα

has order 4 (see Section 3) and m−2α = −1. This gives (a).

It is a standard fact (essentially a calculation in GL(2,R)) that for allh ∈ H, t ∈ Tα, hth

−1 = tsgn(α(h). This proves (b), and (c) follows from thisand the identity α(mβ) = (−1)〈α,β

∨〉. �

The following result is of fundamental importance, and does not hold inthe two root length case (for example for Sp(2n,R)).

Proposition 4.7 Assume G is oddly laced. Then

(4.8) Z(H) = Z(G)H0

We first prove

Lemma 4.9 In the setting of the Proposition we have

(4.10) Z(H) ⊂ Z(G)H0.

Proof. It is enough to show if h ∈ H and {h,mα} = 1 for all real rootsα then h ∈ Z(G)H0. (This is where the oddly laced condition appears:otherwise we only have this identity for the long real roots.) By (4.6)(b) itis enough to show

(4.11) α(h) > 0 for all α ∈ Φr(G,H) implies h ∈ Z(G)H0

This is a straightforward calculation using roots and weights. Choose a basisof the root lattice of the form

α1, . . . , αm, αm+1, . . . , αm+n, αm+n+1, . . . αm+n+2r

20

where αi is real for 1 ≤ i ≤ m, imaginary for m + 1 ≤ i ≤ m + n, andθαm+n+2i−1 = −αm+n+2i for all 1 ≤ i ≤ r. It is a basic fact about latticesthat such a basis exists. Let λ∨1 , . . . , λ

∨m+n+2r be the dual basis of coweights.

For each i choose xi ∈ C so that exi = αi(h). If i ≤ m we may assumexi ∈ R since αi(h) > 0. We may also assume xm+n+2i = xm+n+2i−1 for all1 ≤ i ≤ r. Let X =

∑i xiλ

∨i , h1 = exp(X). It follows easily that X ∈ h,

h1 ∈ H0, and α(h) = α(h1) for all roots α. Let z = hh−11 ; this is contained

in G, and also Z(G(C)), since α(z) = 1 for all roots α. Therefore z ∈ Z(G),and h = zh1 ∈ Z(G)H0. �

Proof of the Proposition. The statement is equivalent to

(4.12) Z0(H) = Z0(G)H0.

It is enough to show Z0(H) ⊂ Z0(G)H0 (the reverse inclusion is obvious). Wefirst prove this assuming H is a maximally split Cartan subgroup. Supposeh ∈ Z0(H). By the Lemma h = zy with z ∈ Z(G), y ∈ H0. Then y ∈ Z0(H),so z = hy−1 ∈ Z0(H). It is enough to show z ∈ Z0(G). We have {z, g} = 1for all g ∈ H , and also for g ∈ G0 by (4.3). Since H is maximally splitG = HG0, so z ∈ Z0(G).

The general case will be proved after Lemma 4.25, using Cayley trans-forms. �

Suppose H is a Cartan subgroup and α is a real or noncompact imaginaryroot. We define the Cayley transform Hα of H with respect to α as in [22,§11.15]; also see [26, Proposition 8.3.4 and 8.3.8]. Then Hα has a noncompactimaginary or real root β, and the Cayley transform of Hα with respect to βis H .

In order to emphasize the symmetry of the situation we change notationand let Hα = TαAα be a Cartan subgroup with a real root α. Then we letHβ = TβAβ be its Cayley transform, with noncompact imaginary root β.

Define Zα ∈ p and Zβ ∈ k as in [26, Proposition 8.3.4 and 8.3.8], and letBα = exp(RZα) ≃ R+ ⊂ Hα, and Bβ = exp(RZβ) ≃ S1 ⊂ Hβ. It followseasily from [26, 8.3.4 and 8.3.13] that we have

(Hα ∩Hβ)Bα ⊂ Hα index 1 or 2(4.13)(a)

(Hα ∩Hβ)Bβ = Hβ.(4.13)(b)

21

The root α takes positive real values on (Hα ∩Hβ)Bα. We say α is type I ifinclusion (a) is an equality. Otherwise α is of type II, in which case Hα isgenerated by the left hand side and an element t satisfying α(t) = −1.

We say β is type I if sβ 6∈ W (G,Hβ), and type II otherwise. Then α, βare both of type I or both of type II.

There is an element g ∈ Gad(C) so that if c = Ad(g), c∗ = Ad∗(g) then

c centralizes hα(C) ∩ hβ(C)(4.14)(a)

c(hβ(C)) = hα(C)(4.14)(b)

c∗(α) = β.(4.14)(c)

Note that any c satisfying (a) and (b) satisfies c∗(α) = ±β.

Definition 4.15 Suppose χα is a character of Hα, and χβ is a character ofHβ. We say χα is a Cayley transform of χβ, and vice-versa, if

χα(h) = χβ(h) (h ∈ Hα ∩Hβ)(4.16)(a)

Ad∗(c)(dχα) = dχβ.(4.16)(b)

where c ∈ Gad(C) satisfies (4.14)(a-c).

Here is a convenient alternative characterization of Cayley transforms,which does not refer to the element c. The proof is elementary.

Lemma 4.17 In the setting of the Definition, χα is a Cayley transform ofχβ, and vice versa, if and only if (4.16)(a) holds, and

(4.18) 〈dχα, α∨〉 = 〈dχβ, β∨〉.

Remark 4.19 In the definition of Cayley transform we are allowed to re-place α and β with their negatives. This does not affect the Cartan sub-groups. Suppose χβ is the Cayley transform of χα with respect to {α, β}.Then the Cayley transform of χα with respect to {α,−β} is sβ(χβ), as isclear from the preceding Lemma. Similar commments apply to α.

Lemma 4.20 Fix α and β.(1) Fix a character χα of Hα. There is a Cayley transform χβ of χα if andonly if

(4.21) 〈dχα, α∨〉 ∈ Z

22

and

(4.22) χα(mα) = (−1)〈dχα,α∨〉.

Assume these hold. Then the Cayley transform of χα is given by (4.16)(a)and

(4.23) χβ(exp(xZβ)) = ei〈dχ,α∨〉x.

(2) Fix a character χβ of Hβ. There are one or two choices of Cayley trans-form χα of χβ, given as follows. Define χα restricted to (Hα ∩ Hβ)Bα by(4.16)(a), and

(4.24) χα(exp(xZα)) = e〈dχβ ,β∨〉x.

If β is of type I this defines the character χα of Hα. If β is of type II defineχ±α to be the two extensions of χα to Hα.

Sketch of proof. This is elementary from the identities [26, 8.3.4 and8.3.13]. See [26, Lemma 8.3.7 and 8.3.15], where the setting is a regularcharacter and the construction differs from this one by a ρ-shift. For thisreason the proof of the Lemma is in fact much easier than those in [26]. �

We now consider Cayley transforms for G in the oddly laced case. Theanalogue of (4.13) is:

Lemma 4.25 Assume G is oddly laced and G is an admissible cover of G.Then

(Z(Hα) ∩ Z(Hβ))Bβ = Z(Hβ)(4.26)(a)

(Z(Hα) ∩ Z(Hβ))Bα = Z(Hα)(4.26)(b)

Proof. We first prove (a). We have to show

(4.27) (Z0(Hα) ∩ Z0(Hβ))Bβ = Z0(Hβ).

Since Bβ is connected, Bβ ⊂ H0β ⊂ Z0(Hβ), and the inclusion ⊂ is clear. For

the opposite inclusion suppose g ∈ Z0(Hβ). By (4.13) write g = hb withh ∈ Hα ∩ Hβ, b ∈ Bβ. Then h = gb−1 ∈ Z0(Hβ) since both g and b are inZ0(Hβ). It is enough to show h ∈ Z0(Hα).

23

Note that {h, x} = 1 for x ∈ Hα ∩ Hβ (since h ∈ Z0(Hβ)). Since Bα

is connected Bα ⊂ H0α ⊂ Z0(Hα), and therefore {h, x} = 1 for x ∈ Bα.

Therefore {h, x} = 1 for x ∈ (Hα ∩Hβ)Bα. If α is type I the right hand sideis Hα and we are done. Otherwise choose t satisfying α(t) = −1, so that Hα

is generated by (Hα ∩Hβ)Bα and t.It is enough to show {h, t} = 1. If this is not the case replace h with hmα

and b with bmα (note thatmα ∈ Hα∩Hβ and Bβ). By (4.6)(b) {mα, t} = −1,so {hmα, t} = 1.

For (b) the inclusion ⊂ is immediate since Bα is connected. On the otherhand suppose g ∈ Hα but g 6∈ (Hα∩Hβ)Bα. Then α(g) < 0, and by (4.6)(b){mα, g} = −1, so g 6∈ Z0(Hα). This proves the reverse inclusion. �

We may now complete the proof of Proposition 4.7.Proof. We have already shown this for the most split Cartan subgroup.By repeated use of the Cayley transform it is enough to show that in theprevious setting

(4.28) Z(Hα) = Z(G)H0α ⇒ Z(Hβ) = Z(G)H0

β

By the Lemma we have

(4.29)

Z(Hβ) = (Z(Hα) ∩ Z(Hβ))Bβ

= (Z(G)H0α ∩ Z(Hβ))Bβ

= Z(G)(H0α ∩ Z(Hβ))Bβ

= Z(G)H0β

For the last equality we have used that (H0α ∩ Hβ)Bβ ⊂ H0

β, which implies

(H0α ∩ Z(Hβ))Bβ ⊂ H0

β. �

We now define Cayley transforms for genuine characters as in Definition4.15, using Z(Hα), Z(Hβ), and Lemma 4.25 in place of (4.13).

Definition 4.30 Assume G is oddly laced and G is an admissible cover ofG. Fix α, β, Hα and Hβ. Suppose χα is a genuine character of Z(Hα), and

χβ is a genuine character of Z(Hβ). We say χα is a Cayley transform of χβ,and vice-versa, if

χα(h) = χβ(h) (h ∈ Z(Hα) ∩ Z(Hβ))(4.31)(a)

Ad∗(c)(dχα) = dχβ.(4.31)(b)

where c ∈ Gad(C) satisfies (4.14)(a-c).

24

Unlike in the linear case this defines χβ uniquely, since (4.26)(a) is anequality, rather than the containment of 4.13(a)).

The analogue of Lemma 4.17 follows easily from Proposition 4.7.

Lemma 4.32 In the setting of the Definition χα and χβ are each othersCayley transforms if and only if

dχα(X) = dχβ(X) (X ∈ hα ∩ hβ),(4.33)(a)

χα(z) = χβ(z) (z ∈ Z(G))(4.33)(b)

〈dχα, α∨〉 = 〈dχβ, β∨〉.(4.33)(c)

The analogue of Lemma 4.20 is:

Lemma 4.34 Suppose we are in the setting of Definition 4.30.(1) Suppose χα is a genuine character of Z(Hα). If α is type II, then theCayley transform χβ of χα exists if and only if

(4.35)(a) 〈dχα, α∨〉 ∈ Z +1

2.

Assume this holds. Then χβ is given by (4.31)(a) and

(4.35)(b) χβ(exp(xZβ)) = e〈deχα,α∨〉ix.

If α is type I let mβ = exp(πZβ) ∈ Z(Hα)∩Z(Hβ). Note that χβ(mβ) = ±i.Then the Cayley transform χβ of χα exists if and only if (4.35)(a) and

(4.35)(c) eiπ〈deχα,α∨〉 = χ(mβ).

In this case it is given by (4.31)(a) and (4.35)(b).

(2) Suppose χβ is a genuine character of Z(Hβ). Then there is a uniqueCayley transform χα of χβ given by (4.31)(a) and

(4.35)(d) χα(exp(xZα)) = e〈deχβ ,β∨〉x.

Proof. For (1) it is immediate from the definition that χβ is defined by

(4.31)(a) and (4.35)(b). If α is type II then Z(Hα) ∩ Z(Hβ) ∩ Bβ = 1, and

there is no further condition. If α is type I then mβ ∈ Z(Hα) ∩Z(Hβ) ∩ Bβ ,which gives condition (4.35)(c). Case (2) is similar, and easier. �

25

5 Transfer Factors: Special Case

Let (G, G,G) be an admissible triple as in Definition 3.3. The basic ideaof transfer factors is simple when G = G is connected and semisimple (forexample take G(C) = G(C) semisimple and simply connected). We discussthis case, before turning to the general case in the next two sections. Westart with some general notation before specializing to the case at hand.

Suppose h = t⊕a is a Cartan subalgebra of g, with corresponding Cartansubgroups H,H and H of G, G and G, respectively. Let H ′, H ′, H

′be the

regular elements of these groups.Let Φ+ be a set of positive roots of H in G. For h ∈ H ′ define

∆0(h,Φ+) =∏

α∈Φ+

(1 − e−α(h))(5.1)(a)

ǫr(h,Φ+) = sign

∏

α∈Φ+r

(1 − e−α(h))(5.1)(b)

∆1(h,Φ+) = ǫr(h,Φ+)∆0(h,Φ+)(5.1)(c)

Let D(h) be the coefficient of trank G in Det(t+ 1 − Ad(h)), and let n bethe number of positive roots. It is easy to see that

(−1)nD(h) = ∆0(h)2e2ρ(h)(5.1)(d)

|D(h)| 12 = |∆0(h)||eρ(h)|(5.1)(e)

where the final term is shorthand for |e2ρ(h)| 12 . It is evident these expressionsare independent of the choice of Φ+.

If G(C) is acceptable define

(5.1)(f) ∆(h,Φ+) = eρ(h)∆0(Φ+, h),

in which case

(5.1)(g) |D(h)| 12 = |∆(h,Φ+)|,

independent of the choice of Φ+.The same definitions apply to G (by pulling back from G) and to G.

Definition 5.2 Let

(5.3)(a) X(G, G) = {(h, g) ∈ G× G |φ(h) = p(g)}.

26

and

(5.3)(b) X(H, H) = X(G, G) ∩H × H.

For g ∈ G let

(5.3)(c) X(G, g) = {h ∈ G |φ(h) = p(g)} = {h ∈ G | (h, g) ∈ X(G, G)}.

and define X(H, g) similarly.

Let X ′(G, G), X ′(H, H) be the subsets consisting of regular semisimpleelements. With G playing the role of G we have φ(h) = h2 and

(5.4) X(H, H) = {(h, g) ∈ H × H | p(g) = h2}.

For (h, g) ∈ X(H, H), choose h satisfying p(h) = h and define

(5.5) τ(h, g) = h2g−1.

This is clearly independent of the choice of h, and τ(h, g) = ±1 since p(h2g−1) =h2p(g−1) = 1.

Now suppose G = G is semisimple and connected. Recall (Definition3.14) we always assume Gd(C) is acceptable. Since G = Gd, ∆ is defined,

and for (h, g) ∈ X(H, H) it is natural to consider

(5.6)∆(h,Φ+)

∆(g,Φ+)Γ0(h, g)

for some factor Γ0(h, g). See the Introduction. Note that the quotient isindependent of the choice of Φ+. We want Γ0 to be a genuine function of g,i.e. Γ0(h,±g) = ±Γ0(h, g). We also want Γ0(h, g) to have small finite orderfor all h, g.

Considerations of harmonic analysis make it natural to include the ǫrterms (cf. 5.1(c)) and instead define

(5.7) ∆(h, g) =∆(h,Φ+)ǫr(h,Φ

+)

∆(g,Φ+)ǫr(g,Φ+)Γ0(h, g)

with Γ0(h, g) to be determined. Now the quotient depends on the choice ofΦ+r .

27

Fix (h, g) ∈ X(H, H). The simplest genuine function on X(H, H) is τ(5.5). Motivated by the case of GL(n) (see the Introduction) it is natural touse this on H0:

(5.8) Γ0(h, g) = τ(h, g) = h2g−1 = ±1 (h ∈ H0).

Here is another way to think of this term. Let χ be a genuine character of

H0. Then χ2 factors to a character of H0: for h ∈ H0 define χ2(h) = χ(h2)

where p(h) = h. Then it is easy to see that

(5.9) Γ0(h, g) =χ(g)

χ2(h)(h ∈ H0).

This suggests a way to define Γ0(h, g) for all h; choose a character χ0 ofH restricting to χ2 on H0, and define:

(5.10) Γ0(h, g) =χ(g)

χ0(h)for all (h, g) ∈ X(H, H).

Recall since G is connected by (2.7) H = Γr(H)H0. A basic property ofχ is the following (cf. 6.21):

(5.11) χ2(t) = eρr(t) t ∈ Γr(H) ∩H0

where ρr = ρ(Φ+r ). Note that eρr only depends on the positive real roots,

and its restriction to H0 is independent of all choices, since any two differ bya sum of real roots, which is necessarily trivial on Γr(H) ∩H0.

Therefore we can define χ0 by

(5.12) χ0(h) =

{χ2(h) h ∈ H0

eρr(h) h ∈ Γr(H).

Note that χ0 depends on the choice of Φ+, although its restriction to Γr(H)∩H0 does not. Defining Γ0(h, g) by (5.10), we see:

Lemma 5.13 Fix a set of positive roots Φ+, and let ρr = ρ(Φ+r ). There is a

unique function Γ0 on X(H, H) satisfying

(5.14)Γ0(h, g) = τ(h, g) (h ∈ H0)

Γ0(th, g) = e−ρr(t)Γ0(h, g) (t ∈ Γr(H), h ∈ H).

28

Example 5.15 Let G = SL(2,R) and let H be the diagonal Cartan sub-group, which we identify with R×. If p(x) = x and ǫ = ±1 then (x, x2ǫ) ∈X(H, H) and

(5.16) Γ0(x, x2ǫ) = sgn(x)ǫ,

independent of Φ+.

While this is a natural extension of Γ0(h, g) from H0 to H , the compellingevidence that it is the correct definition amounts to the matching conditionsinvolving different Cartan subgroups. See Lemma 6.37.

Definition 5.17 Assume G = G is connected and semisimple. Fix a set ofpositive roots Φ+. For (h, g) ∈ X(H, H) define the transfer factor

(5.18) ∆(h, g) =∆(h,Φ+)ǫr(h,Φ

+)

∆(g,Φ+)ǫr(g,Φ+).Γ0(h, g)

While Γ0(h, g) depends on a choice of Φ+ (actually only Φ+r ), the transfer

factor itself is independent of this choice:

Lemma 5.19 ∆(h, g) is independent of the choice of Φ+.

Proof. Write h = th0 with t ∈ Γr(H), h0 ∈ H0. By (5.14)

(5.20) ∆(h, g) =∆(h,Φ+)

∆(g,Φ+)

ǫr(h,Φ+)

ǫr(g,Φ+)eρr(Φ+)(t)τ(h0, g)

Let Φ+r (h0) = {α ∈ Φr | eα(h0) > 1}. By (7.6)(d) this gives

(5.21) ∆(h, g) =∆(h,Φ+)

∆(g,Φ+)eρ(Φ

+r (h0))(t)τ(h0, g)

which is obviously independent of Φ+. �

Lemma 5.22 For all (h, g) ∈ X(G, G) we have

(5.23) ∆(h, g) = c(h, g)|D(h)| 12|D(g)| 12

29

and

(5.24) ∆(h, g) = c′(h, g)∆(h,Φ+)

∆(g,Φ+)

where c(h, g)4 = c′(h, g)2 = 1.

Suppose x ∈ G and let x = p(x). Then

(5.25) ∆(xhx−1, xgx−1) = ∆(h, g).

This is immediate.In the next section we will drop the assumption that G is acceptable, in

which case ∆(h,Φ+) and ∆(g,Φ+) are not necessarily well defined. Withthis in mind we rewrite (5.18) as follows:

(5.26)

∆(h, g) =∆(h,Φ+)ǫr(h,Φ

+)

∆(g,Φ+)ǫr(g,Φ+)

χ(g)

χ0(h)

=∆0(Φ+, h)ǫr(h,Φ

+)

∆0(Φ+, g)ǫr(g,Φ+)

χ(g)eρ(h)

χ0(h)eρ(g)

=∆1(Φ+, h)

∆1(Φ+, g)

χ(g)

(χ0eρ)(h)

since p(g) = h2. Let χ = χ0eρ and assume Φ+ is special (Definition 2.1).

Then by (2.4)(b) and (5.12) for h ∈ Γr(H) we have

(5.27) χ(h) = eρr(h)eρ(h) = ζcx(h).

Thus χ is defined by

(5.28)(a) χ(h) =

{(χ2eρ)(h) h ∈ H0

ζcx(h) h ∈ Γr(H).

Define

(5.28)(b) Γ(h, g) = χ(g)/χ(h),

and then (still assuming G = G is connected and semisimple)

(5.28)(c) ∆(h, g) =∆1(Φ+, h)

∆1(Φ+, g)Γ(h, g).

30

This has the advantage that the quotient is defined in general and thisis the starting point of the definition of transfer factors. There are severalissues in extending it to the general case. First of all eρ is only defined onH0d (recall Gd is acceptable), so (5.28) only defines χ, and hence Γ and ∆,

on Γr(H)H0d = H ∩ G0

d (cf. (2.7)). This may be a proper subgroup of H .Furthermore if G 6= G then some additional modification is required. Wetake up these issues in the next two sections.

6 Characters of Cartan Subgroups.

Let (G, G,G) be an admissible triple. In the previous section we defined thefactor Γ(h, g) (5.28)(b) if G = G is connected and semisimple. In this sectionwe consider the general case, which we use in the subsequent section to definetransfer factors in general.

Suppose H is a Cartan subgroup of G and let Φ+ be a set of positiveroots. Suppose χ is a genuine character of Z(H). Since Gd(C) is acceptableρ exponentiates to a character of (H∩Gd)

0, and χ2eρ is defined on (H∩Gd)0.

Let H be the corresponding Cartan subgroup of G and set Hd = H ∩ Gd.We now use property (3c) of Definition 3.14, which says that χ2eρ factors to

H0

d. The canonical character ζcx of Γ(H) was defined in Section 2.We use (5.28) as our starting point.

Definition 6.1 Let H be a Cartan subgroup of G. Choose

1. a special set of positive roots Φ+ (Definition 2.1)

2. a genuine character χ of Z(H),

3. a character χ of H.

Assume these satisfy:

χ(h) = (χ2eρ)(h), (h ∈ H0

d)(6.2)(a)

χ(h) = ζcx(h) (h ∈ Γr(H)).(6.2)(b)

Fix (Φ+, χ, χ) satisfying these conditions. Suppose (h, g) ∈ X(H, H) (cf.(5.3)(b)). Then we define

(6.3) Γ(χ, χ)(h, g) =χ(g)

χ(h).

31

Let

S(H,Φ+) = {(χ, χ)} satisfying (6.2)(a) and (b)},(6.4)(a)

T (H,Φ+) = {Γ(χ, χ) | (χ, χ) ∈ S(H,Φ+)}(6.4)(b)

S(H) = ∪Φ+S(H,Φ+)(6.4)(c)

T (H) = ∪Φ+T (H,Φ+).(6.4)(d)

(The unions are over special sets of positive roots).

Example 6.5 Let G = G = SL(2,R) and let H be the diagonal Cartansubgroup, which we identify with R×. Let Φ+ = {α} where α(x) = x2. Ifp(x) = x and ǫ = ±1 then a short calculation gives

(6.6) Γ(x, x2ǫ) = x−1ǫ.

This only depends on Φ+; the other choice of Φ+ gives xǫ. Compare Example5.15.

We begin with some elementary properties. Fix H , (χ, χ) as above andlet Γ = Γ(χ, χ). First of all we obtain a character λ of Z0(H).

Lemma 6.7 Let λ = (χ2/χ), viewed as a character of Z0(H), i.e. λ(h) =

χ2(h)/χ(p(h)) where p(h) = h. Then

(6.8)(a) λ(h) = e−ρ(h) (h ∈ Z0(H) ∩Gd).

Also let

(6.8)(b) µ = −dλ− ρ = dχ− 2dχ− ρ.

Then µ|hd= 0 and we identify it with an element of z(C)∗; as such it is the

negative of the differential of λ|Z(G)0.

Furthermore for all (h, g) ∈ X(H, H)

(6.8)(c) Γ(h, g)2 = λ(φ(h)).

In particular if h ∈ Gd then Γ(h, g)2 = e−2ρ(h).

Proof. If h ∈ H0d then by 6.2(b) χ(p(h)) = (χ2eρ)(h). The same identity

holds for h ∈ Γr(H) ∩ Z0(H) by (6.21). In both cases we conclude λ(h) =

32

χ2(h)/χ(p(h)) = e−ρ(h). This holds on Γr(H)H0d ∩ Z0(H), which equals

Z0(H) ∩Gd by (2.7). This gives (a).

For (b), if (h, g) ∈ X(H, H) then Γ(h, g)2 = χ2(g)/χ(h2). Since p(g) =φ(h) and p(φ(h)) = h2 this equals χ2(φ(h))/χ(φ(h)) = λ(φ(h)). The finalequality follows from (a) and (b) and the fact that p(φ(h)) = h2. �

Recall (5.5) for (h, g) ∈ X(H, H), τ(h, g) = h2g−1 = ±1.

Lemma 6.9 Suppose (h, g) ∈ X(H, H).(1) Assume h ∈ p(Z0(H)), and choose y ∈ Z0(H) satisfying p(y) = h. Thenwith λ as in (6.36)(c),

(6.10) Γ(h, g) = λ(y)τ(y, g).

(2) If h ∈ Gd write h = th0 with t ∈ Γr(H), h0 ∈ H0

d, and choose y0 ∈ H0

satisfying p(y0) = h0. Then

(6.11) Γ(h, g) = ζcx(t)e−ρ(y0)τ(y0, g),

and this is independent of all choices except that of Φ+.

(3) In general write h = th0 with t ∈ Γ(H) and h0 ∈ H0

(2.7). Writeg = γg0 where p(γ) = φ(t) and p(g0) = φ(h0). Finally choose y0 ∈ H0

satisfying p(y0) = h0. Then

(6.12) Γ(h, g) =χ(γ)

χ(t)λ(y0)τ(y0, g0).

Note that χ(γ)4 = 1 and χ(t)2 = 1.

Proof. For (1) choose y ∈ p−1(y) and write g = y2τ(y, g) (cf. 5.5). ThenΓ(h, g) = χ(y2)τ(y, g)/χ(h) = (χ2(y)/χ(p(y)))τ(y, g) = λ(y)τ(y, g). Part(2) follows from this, (6.2)(b) and (6.8)(a). Part (3) is similar, noting thatΓ(h, g) = Γ(t, γ)Γ(h0, g0). �

Example 6.13 Let G = G = GL(n,R) and suppose H is the diagonal(split) Cartan subgroup. For simplicity we assume n is even, in which caseZ0(H) = H0. Note that Γ(H) = {diag(ǫ1, . . . , ǫn)} with ǫi = ±1, and Γr(H)is the subset of Γ(H) of elements of determinant 1. If h ∈ H write h = th0

with t ∈ Γ(H) and h ∈ H0. Let Φ+ be any (necessarily special) set of positiveroots.

33

The cover Z(H) → Z0(H) splits; let χ be the unique, genuine quadratic

character of Z(H). Note that ρ exponentiates to H0, and we can defineχ(th0) = eρ(h0). Then (χ, χ) ∈ S(H,Φ+) and by (6.12) we have

(6.14) Γ(h, g) = e−ρ(h0)τ(h0, g).

Given χ, the only other allowed choice of χ and Γ is obtained by multi-plying by ν(det(h)) for ν a character of R× (see Lemma 6.19).

Lemma 6.15 Fix a special set of positive roots Φ+.(1) Suppose ν is a character of Z0(H) and (χ, χ) ∈ S(H,Φ+). Then (χ(ν ◦p), χ(ν ◦ φ)) ∈ S(H,Φ+) and

(6.16) Γ(χ(ν ◦ p), χ(ν ◦ φ)) = Γ(χ, χ).

(2) Suppose Φ+1 is another special set of positive roots. There exists a char-

acter τ of H such that for all (χ, χ) ∈ S(H,Φ+), (χ, χτ) ∈ S(H,Φ+1 ), and

(6.17) Γ(χ, χ)(g, h) = Γ(χ, χτ)(g, h)

for all (h, g) ∈ X(H, H) satisfying h ∈ Γ(H)Z(G).

Proof. The first part is straightforward. For the second define τ(h) =

eρ(Φ+1 )−ρ(Φ+)(h) for h ∈ H

0

d, and τ(h) = 1 for h ∈ Γ(H)Z(G). It is easy to see

that the fact that both Φ+ and Φ+1 are special implies eρ(Φ

+1 )−ρ(Φ+)(h) = 1 for

all h ∈ Γ(H)Z(G) ∩H0

d, so τ is well defined. It follows easily that it has thedesired properties. �

There is a natural action of the characters of G on T (H,Φ+): if ψ is aone-dimensional representation of G define

(6.18) ψ · Γ(χ, χ) = Γ(χ, χψ|H).

Lemma 6.19 The space T (H,Φ+) is non-empty, and the action of the groupof characters of G on T (H,Φ+) is transitive. If H is maximally split it issimply transitive.

Proof. Fix any genuine character χ of Z(H). For existence it is enough toshow that for any special set of positive roots,

(6.20) ζcx(h) = eρχ2(h) h ∈ Γr(H) ∩H0

d.

34

Note that p−1(H0

d) ⊂ p−1(H0) = H0C ⊂ H0Z0(G) = Z0(H) by 3.14(3b))

and Proposition 4.7. Pulling back to G it is therefore enough to show

(6.21) ζcx(h) = eρχ2(h) h ∈ Γr(H) ∩ Z0(H).

Lemma 6.22 χ2(h) = eρr(h) for all h ∈ Z0(H) ∩ Γr(H).

Proof. Fix h ∈ Z0(H) ∩ Γr(H). By (2.6) write h =∏n

i=1mi where{α1, . . . , αn} is a subset of the simple roots for Φ+

r , and mi = mαi. Then

eρr(h) = (−1)n.

For each i choose an inverse image mi of mi in H . By Lemma 3.2 andthe discussion preceding it m2

i = −1 for all i. Assume for the moment thatmi, mj commute for all i, j. Then

(6.23)

χ2(h) = χ([m1 . . . mn]2)

= χ(m21 . . . m

2n)

= χ((−1)n) = (−1)n

since χ is genuine, proving the result.Thus it is enough to show the mi commute. Using the commutator no-

tation of Section 4, the fact that {mi, h} = 1 implies by (4.6)(c) that, foreach i, αi is not orthogonal to an even number of other αj . Consider thesubdiagram of the Dynkin diagram of Φr consisting of the αi (1 ≤ i ≤ n):every node is adjacent to an even number of other nodes. This implies thenodes are all orthogonal, and the mi commute. �

Therefore

(6.24) χ2eρ(h) = eρr(h)eρr+ρcx+ρi(h) = eρcx(h)

since e2ρr(h) = eρi(h) = 1. This equals ζcx(h) by (2.4)(b).Now suppose Γ1,Γ2 ∈ T (H,Φ+). If h ∈ H then ψ0(h) = Γ1(h, g)/Γ2(h, g)

is independent of the choice of g so that (h, g) ∈ X(H, H), and is a characterof H. By (6.9) ψ0(h) = 1 for h ∈ Gd.

We have a surjection

(6.25) Hom(G/Gd,C) → Hom(H/Hd,C)

dual to the injection H/Hd → G/Gd, which is an isomorphism if H is maxi-mally split. If we choose any preimage ψ of ψ0 in (6.25) then it is easy to see

35

that Γ1 = ψΓ2. It follows that characters of G act transitively on T (H,Φ+),and this action is simply transitive if H is maximally split. �

We now need to choose elements of T (H,Φ+) consistently for all H . Wedo this by reducing to the most split Cartan subgroup. For this we need aLemma.

Lemma 6.26 Suppose (χ, χ) ∈ S(H,Φ+). Then

(6.27) χ(h) = χ(wh) (w ∈W (G,H), h ∈ Γ(H)Z(G)).

Proof. We need to show χ(g(wg)−1) = 1 for g ∈ Γ(H). This is obviousif w ∈ W (Φi). If w ∈ W (Φr) then g(wg)−1 ∈ Γr(H). Then χ(g(wg)−1) =ζcx(g(wg)

−1) = 1 by (6.2)(b) and (2.4)(a).By [27, Proposition 3.12] W (G(C), H(C))θ, which contains W (G,H), is

generated by W (Φi),W (Φr) and elements of the form sαsθα where 〈α, θα∨〉 =0 and sαsθαΦ

+r = Φ+

r . So it is enough to show χ(g(wg)−1) = 1 for w of thisform.

Write g = expπiX with X ∈ a. Since g2 = 1, α(X) ∈ Z. Then

(6.28) g(wg)−1 = expπiα(X)(α∨ + θα∨) ∈ Γ(H) ∩H0

d

Let ρ = ρ(Φ+). Then

(6.29)χ((wg)g−1) = (χ2eρ)(g(wg)−1) (by (6.2)(a))

= (−1)α(X)〈2deχ,α∨+θα∨〉eρ−w−1ρ(g)

By [3, Lemma 6.11] 〈dχ, α∨ + θα∨〉 ∈ Z. The fact that Φ+ is special andwΦ+

r = Φ+r implies ρ − w−1ρ is a sum of imaginary roots and terms β − θβ

with β complex. Therefore ew−1ρ−ρ(g) = 1. �

Fix a maximally split Cartan subgroup Hs = TsAs of G and (χs, χs) ∈S(Hs).

Suppose H is a Cartan subgroup, and write H = TA as usual. Thereexists x ∈ G such that xAx−1 ⊂ As, and therefore xΓ(H)x−1 ⊂ Γ(Hs).

Suppose (h, g) ∈ X(H, H) where h ∈ Γ(H)Z(G). By Lemmas 3.11(5) and

3.12 φ(h) ∈ Z0(G) so g ∈ Z(G) ⊂ Z(Hs). Therefore (xhx−1, g) ∈ X(Hs, Hs).Define

(6.30) Γs(h, g) = Γ(χs, χs)(xhx−1, g).

By the Lemma this is independent of the choice of x.

36

Proposition 6.31 Fix a maximally split Cartan subgroup Hs and (χs, χs) ∈S(Hs). Suppose H is a Cartan subgroup, χ is a genuine character of Z(H),and Φ+ is a special set of positive roots for H. Then there exists a uniquecharacter χ of H such that (χ, χ) ∈ S(H,Φ+) and

(6.32) Γ(χ, χ)(h, g) = Γs(h, g)

for all (h, g) ∈ X(H, H) with h ∈ Γ(H)Z(G).

Explicitly (6.32) is

(6.33)χ(g)

χ(h)=

χs(g)

χs(xhx−1)

or alternatively

(6.34) χ(h) = (χ/χs)(φ(h))χs(xhx−1).

Before giving the proof of the Proposition we give the main definition ofthis section.

Definition 6.35 Fix a split Cartan subgroup Hs and (χs, χs) ∈ S(Hs). Sup-pose H is a Cartan subgroup and Φ+ is a special set of positive roots.

(1) Let S(H,Φ+, χs, χs) be the set of pairs (χ, χ) ∈ S(H,Φ+) satisfying(6.32).

(2) Suppose χ is any genuine character of H. Choose χ so that (χ, χ) ∈S(H,Φ+, χs, χs) and let

(6.36)(a) Γ(H,Φ+) = Γ(χ, χ) ∈ T (H,Φ+).

Also define

(6.36)(b) λ(H,Φ+) = χ2/χ ∈ Z0(H)

and

(6.36)(c) µ = −dλ− ρ = dχ− 2dχ− ρ ∈ z(C)∗

(cf. Lemma 6.7). If it is necessary to indicate the dependence on (χs, χs) wewrite Γ(χs, χs, H,Φ

+), λ(χs, χs, H,Φ+) and µ(χs, χs).

37

It follows easily from Lemma 6.15 that Γ(H,Φ+) and λ(H,Φ+) are indepen-dent of the choice of χ. It is also easy to see that µ = dχs − 2dχs restrictedto z(C)∗, and is therefore independent of the choice of (H,Φ+).

We now prove Proposition 6.31. Using Cayley transforms (Section 4)we will reduce to the following Lemma. Suppose Hα is a Cartan subgroup,with real root α, and Hβ is the Cayley transform of Hα, with noncompactimaginary root β. See the discussion preceding (4.13). Fix c ∈ Aut(g)satisfying (4.14)(a-c).

Lemma 6.37 Suppose Φ+α is a special set of positive roots for Hα and (χα, χα) ∈

S(Hα,Φ+α ). Suppose Φ+

β is a special set of positive roots for Hβ, and χβ is a

genuine character of Z(Hβ). Then there exists a unique character χβ of Hβ

such that (χβ, χβ) ∈ S(Hβ,Φ+β ) and

(6.38) Γ(χα, χα)(h, g) = Γ(χβ, χβ)(h, g)

for all (h, g) ∈ X(Hβ, Hβ) satisfying h ∈ Γ(Hβ)Z(G).In addition assume c∗(Φ+

α ) = Φ+β . Then (6.38) holds for all h ∈ Hα∩Hβ.

Proof. To avoid runaway notation let H = Hβ for a moment. The character

χβ is determined on Γ(H)Z(G) by (6.38) and on H0

d by (6.2)(a). Since

H0 ⊂ Z(G)H

0

d, by (2.7) H = Γ(H)Z(G)H0

d, and uniqueness is immediate.For existence, by Lemma 6.15 it is enough to prove this for a single

choice of Φ+α and Φ+

β . It is not hard to see we can choose Φ+α special so that

Φ+β = c∗(Φ+

α ) is also special. This implies α is simple for Φ+r (cf. Lemma

14.17).Fix χβ. By 6.15(1) we may choose χα arbitrarily, so by Lemma 4.34

choose χα to be the Cayley transform of χβ. Since χβ is genuine and βis a noncompact imaginary root, 〈dχβ, β∨〉 ∈ Z + 1

2by [3, Lemma 6.11].

Therefore 〈dχα, α∨〉 = 〈dχβ, β∨〉 ∈ Z + 12. Let χα be any character of Hα

with (χα, χα) ∈ S(Hα,Φ+α ). Then

(6.39) χα(mα) = ζcx(mα) = (−1)〈ρ−ρr ,α∨〉 = −(−1)〈ρ,α∨〉

since α is simple for Φ+r . Furthermore

(6.40) 〈dχα, α∨〉 = 〈2dχα + ρ, α∨〉 ∈ 2Z + 1 + 〈ρ, α∨〉.

Thus χα satisfies the conditions of Lemma 4.20. Let χβ be the (unique)Cayley transform of χα.

38

It is enough to show (χβ, χβ) ∈ S(Hβ,Φ+β ), for then it is obvious from the

definition of Cayley transforms that (6.38) holds for all h ∈ Hα ∩Hβ.We have

(6.41)

dχβ = Ad∗(c)(dχα)

= Ad∗(c)(2dχα + ρ(Φ+α ))

= 2dχβ + ρ(Φ+β )

by the choice of Φ+α and Φ+

β . This verifies (6.2)(a).

We now verify (6.2)(b). Suppose γ is a real root of Hβ. Then mγ ∈Hα ∩Hβ, and χβ(mγ) = χα(mγ) = ζcx(G,Hα)(mγ), by (6.2)(b) for (χα, χα).We want to show this equals ζcx(G,Hβ)(mγ), i.e.

(6.42) ζcx(G,Hα)(mγ) = ζcx(G,Hβ)(mγ).

If the simple factor containing α, β is of type G2 an explicit calculationshows that both sides are 1, so assume G is simply laced.

The root γ = Ad∗(c−1)(γ) of Hα is also real, and 〈γ, α∨〉 = 0. Supposeδ is a complex root of Hα with 〈δ, γ∨〉 6= 0. Then the corresponding rootof Hβ is also complex. Conversely, suppose δ is a complex root of Hβ with

〈δ, γ∨〉 6= 0. Then the corresponding root δ of Hα is complex unless sβδ = σδ,in which case it is real. In this case, let δ′ = sγδ. It is another complex root

of Hβ with δ′ real. Since 〈δ, γ∨〉 6= 0, δ′ 6= δ, and since γ is real, δ′ 6= σδ.Since δ 6= ±γ, δ′ 6= −δ. Suppose δ′ = −σδ = −sβδ. Then

(6.43) 2δ = 〈δ, γ∨〉γ + 〈δ, β∨〉β.

This cannot happen since in the simply laced case, two orthogonal roots haveno other roots in their (real) span.

Therefore both δ and sγ(δ) contribute to ζcx(G,Hβ)(mγ), and their totalcontribution is δ(mγ)(sγδ)(mγ) = δ(mγ)δ(mγ) = 1.

It follows that the terms in ζcx(G,Hα) and ζcx(G,Hβ) are the same, withthe exception of those just discussed, which are 1. �

Remark 6.44 Note that Γ(χα, χα)(mα, 1) = χα(mα) and Γ(χβ, χβ)(mα, 1) =χβ(mα). As in the proof of the Lemma χβ(mα) = −(−1)〈ρ,α

∨〉. The equalityof the Lemma then implies χα(mα) = ζcx(mα). This motivates condition(6.2)(b).

39

Proof of the Proposition. This is now straightforward. If H is maximallysplit there is nothing to prove. If H is obtained from Hs by a series of Cayleytransforms we conclude the result by a repeated application of Lemma 6.37.It is easy to check that if the conditions of the Proposition hold for a Cartansubgroup H , they hold for every G-conjugate of H . Up to conjugacy everyCartan subgroup is conjugate to one obtained by a series of Cayley transformsfrom Hs, and the result follows. �

For later use we note a consequence of the last part of the proof of Lemma6.37.

Lemma 6.45 Suppose H1 = T1A1 and H2 = T2A2 are Cartan subgroupswith A1 ⊂ A2. Suppose α ∈ Φr(G,H1).(1)

ζcx(G,H1)(mα) = ζcx(G,H2)(mα).

(2) Let M1 = CentG(A1) and suppose α ∈ Φr(M1, H2). Then

ζcx(G,H2)(mα) = ζcx(M1, H2)(mα).

Proof. The first equality follows from a repeated application of (6.42). Thesecond is similar. Choose a set S of complex roots such that

(6.46) {β ∈ Φcx(G,H2)\Φcx(M1, H2) | 〈β, α∨〉 6= 0} = {±β,±σβ | β ∈ S}.If β ∈ S then sαβ is also complex, is not contained in Φ(M1, H2), and is notequal to ±β,±σβ Then S is can be written as a union over pairs {β, sαβ}and the result follows as in the proof of (6.42). �

The dependence of Γ(H,Φ+) on Φ+ follows easily from Lemma 6.15(2)and its proof.

Lemma 6.47 Suppose Φ+1 ,Φ

+2 are special sets of positive roots for H. Let

(h, g) ∈ X(H, H), and write h = γh0 with γ ∈ Γ(H) and h0 ∈ H0. Then

(6.48) Γ(H,Φ+1 )(h, g) = Γ(H,Φ+

2 )(h, g)eρ(Φ+2 )−ρ(Φ+

1 )(h0).

The factors Γ(H,Φ+) are conjugation invariant in the following sense.

Lemma 6.49 Suppose Φ+ is a special set of positive roots forH, and (h, g) ∈X(H, H). Fix x ∈ G and let x = p(p(x)) ∈ G. Let H1 = xHx−1 andΦ+

1 = Ad(x)Φ+. Then

(6.50) Γ(H1,Φ+1 )(xhx−1, xgx−1) = Γ(H,Φ+)(h, g)

40

Proof. Choose (χ, χ) ∈ S(H,Φ+, χs, χs). Let χ1 = xχx−1 and χ1 = xχx−1.It is clearly enough to show (χ1, χ1) ∈ S(H1,Φ

+1 , χs, χs). Condition (6.2)(b)

is immediate, and for all h ∈ (H1 ∩Gd)0

χ1(h) = χ(x−1tx) = χ2eρ(Φ+)(x−1tx) = χ2

1eρ(Φ+

1 )(h)

so (6.2)(a) holds as well.Assume γ ∈ Γ(H1)Z(G) and choose y satisfying yΓ(H)y−1 ⊂ Γ(Hs). By

(6.34)

(6.51)χ1(γ) = χ(x−1γx) = (χ/χs)(φ(x−1γx))χs(yx

−1γxy−1)

= (χ1/χs)(φ(γ))χs((yx−1)γ(yx−1)−1).

This proves (6.34) holds for (χ1, χ1). �

We summarize the important properties of Γ(H,Φ+), reformulating themslightly.

Proposition 6.52 Fix a split Cartan subgroup, a special set Φ+s of positive

roots, and (χs, χs) ∈ S(Hs,Φ+s ). Suppose H is any Cartan subgroup, Φ+ is

special, and (h, g) ∈ X(H, H).(1)

(6.53)(a) Γ(H,Φ+)(h, g) = e−ρ(y)τ(y, g) (h ∈ H0

d).

(2) Suppose x ∈ G, x = p(p(x)). Then

(6.53)(b) Γ(xHx−1,Ad(x)Φ+)(xhx−1, xgx−1) = Γ(H,Φ+)(h, g)

(3) Suppose (i = 1, 2) (hi, gi) ∈ X(H, H). Then

(6.53)(c) Γ(H,Φ+)(h1h2, g1g2) = Γ(H,Φ+)(h1, g1)Γ(H,Φ+)(h2, g2).

(4) For i = 1, 2 let Hi be a Cartan subgroup and Φ+i a special set of positive

roots for Hi. Assume (h, g) ∈ X(H i, Hi) for i = 1, 2. Choose c ∈ Gad(C)satisfying Ad(c)h2 = h1, Ad(c)|h1∩h2

= 1, and Ad∗(c)(Φ+1 ) = Φ+

2 . Then

(6.53)(d) Γ(H1,Φ+1 )(h, g) = Γ(H2,Φ

+2 )(h, g).

Conditions (1,2,4) and Condition (3) for h1 ∈ Γ(H)Z(G), g1 ∈ Z(G)uniquely determine the functions Γ(H,Φ+), given the restriction of Γ(Hs,Φ

+s )

to (Γ(Hs)Z(G) × Z(G)) ∩X(Hs, Hs).

41

7 Transfer Factors

We now define transfer factors for a general admissible triple (G, G,G), gen-eralizing Definition 5.17.

Definition 7.1 A set of lifting data for (G, G,G) is a pair (χs, χs) ∈ S(Hs)for some maximally split Cartan subgroup Hs of G.

The transfer factor ∆eGG

associated to (χs, χs) is the function on X ′(G, G)(Definition 5.2) defined as follows.

Suppose (h, g) ∈ X ′(G, G). Let H = CentG(p(g)); this is a Cartan sub-group of G. Choose a special set of positive roots Φ+ for H. Recall Γ(H,Φ+)is given by Definition 6.35, and ∆1 is given in (5.1)(c). Define the transferfactors

(7.2) ∆eGG(h, g) =

∆1(Φ+, h)

∆1(Φ+, g)Γ(H,Φ+)(h, g).

Proposition 7.3 ∆eGG

is well defined, i.e. independent of the choice of Φ+.

Proof. We have to show that ∆1(Φ+,h)∆1(Φ+,eg)

and Γ(H,Φ+) satisfy inverse trans-

formation properties with respect to Φ+.Suppose (h, g) ∈ X ′(H, H). By (2.7) write h = γh0 with γ ∈ Γ(H) and

h0 ∈ H0. Suppose Φ+

1 ,Φ+2 are two choices of special positive roots. By

Lemma 6.47

(7.4)(a) Γ(H,Φ+1 )(h, g) = Γ(H,Φ+

2 )(h, g)eρ(Φ+2 )−ρ(Φ+

1 )(h0).

By the definition of ∆eGG

it is enough to prove

(7.4)(b)∆1(Φ+

1 , h)

∆1(Φ+1 , g)

=∆1(Φ+

2 , h)

∆1(Φ+2 , g)

eρ(Φ+1 )−ρ(Φ+

2 )(h0).

Every term factors to G, so we can replace g with p(p(g)) = p(φ(h)) = h2.This reduces us to showing

(7.5)∆1(Φ+

1 , h)

∆1(Φ+1 , h

2)=

∆1(Φ+2 , h)

∆1(Φ+2 , h

2)eρ(Φ

+1 )−ρ(Φ+

2 )(h0).

42

We first compute for any set of positive roots Φ+:

(7.6)(a)

ǫr(h,Φ+)

ǫr(h2,Φ+)= sgn

∏α∈Φ+

r(1 − e−α(h))∏

α∈Φ+r(1 − e−α(h2))

= sgn∏

α∈Φ+r

(1 + e−α(γ)e−α(h0))

A given term is positive unless eα(γ) = −1 and eα(h0) < 1. Let

(7.6)(b) Φ+r (h) = {α ∈ Φr | eα(h0) > 1}.

Then

(7.6)(c)ǫr(h,Φ

+)

ǫr(h2,Φ+)=

∏

α∈Φ+r ∩(−Φ+

r (h))

eα(γ) = eρr(Φ+)−ρ(Φ+r (h))(γ).

Letting Φ+r,j = Φ+

j ∩ Φr we have

(7.6)(d)ǫr(h,Φ

+1 )

ǫr(h2,Φ+1 )

= eρ(Φ+r,1)−ρ(Φ+

r,2)(γ)ǫr(h,Φ

+2 )

ǫr(h2,Φ+2 ).

Now consider the ∆0 term of (5.1)(c). Write Φ+1 = wΦ+

2 for some w ∈W (Φ), so that

(7.6)(e)

∆0(Φ+1 , h)

∆0(Φ+1 , h

2)=

ǫ(w)eρ(Φ+2 )−ρ(Φ+

1 )(h)∆0(Φ+2 , h)

ǫ(w)eρ(Φ+2 )−ρ(Φ+

1 )(h2)∆0(Φ+2 , h

2)

= eρ(Φ+1 )−ρ(Φ+

2 )(h)∆0(Φ+

2 , h)

∆0(Φ+2 , h

2)

since ρ(Φ+1 ) − ρ(Φ+

2 ) is a sum of roots.Now

(7.6)(f)

eρ(Φ+r,1)−ρ(Φ+

r,2)(γ)eρ(Φ+1 )−ρ(Φ+

2 )(h)

= eρ(Φ+r,1)−ρ(Φ+

r,2)−ρ(Φ+2 )+ρ(Φ+

1 )(γ)e−ρ(Φ+2 )+ρ(Φ+

1 )(h0)

= eρ(Φ+1 )−ρ(Φ+

2 )(h0)

since γ has order 2 and (by (2.4)(b))

(7.6)(g) eρ(Φ+1 )−ρ(Φ+

r,1)(γ) = ζcx(G,H)(γ) = eρ(Φ+2 )−ρ(Φ+

r,2)(γ).

Multiplying (c) and (e), and using (f) shows (7.5), and completes the proof.�

Here are some elementary properties of transfer factors.

43

Lemma 7.7 Suppose (h, g) ∈ X(H, H). Fix a special set of positive rootsΦ+ of H, and let ρ = ρ(Φ+), λ = λ(H,Φ+) (cf. 6.36)(b). Let g = p(g). Then

(7.8)(a) ∆(h, g)2 =D(h)

D(g)e2ρ(h)λ(φ(h)).

Write h = th0 with t ∈ Γ(H)Z(G) and h0 ∈ H0

d. Then

(7.8)(b) ∆(h, g)2 =D(h)

D(g)e2ρ(t)λ(φ(t)).

In particular

(7.8)(c) ∆(h, g) = c(h, g)λ(φ(t))12|D(h)| 12|D(g)| 12

where c(h, g)4 = 1. Finally if λ is unitary then

(7.8)(d) |∆(h, g)| =|D(h)| 12|D(g)| 12

.

Proof. The main point is that

(7.9)

[∆0(Φ+, h)ǫr(h,Φ

+)

∆0(Φ+, g)ǫr(g,Φ+)

]2

=∆0(Φ+, h)2e2ρ(h)

∆0(Φ+, g)2e2ρ(g)

e2ρ(g)

e2ρ(h)=D(h)

D(g)e2ρ(h),

the last equality follows from (5.1)(d) and the fact p(p(g)) = h2. Then(7.8)(a) and (7.8)(b) follow from (6.36)(c), and (c) and (d) are elementaryconsequences of this. �

Remark 7.10 Recall µ ∈ z(C)∗ is given in Definition 6.35. If µ = 0(i.e. λ|Z(G)0 = 1) then in (7.8)(c) λ(φ(t))2 = 1, and ∆(h, g) differs from

|D(h)/D(g)| 12 by a fourth root of unity. By varying our choice of liftingdata (χs, χs) we are free to choose the restriction of µ to Z(G)0, up tothe constraints (6.8)(a). For simplicity assume G = G. It is easy to seewe can take this restriction to be trivial if and only if eρ(z) = 1 for allz ∈ Z(G)0 ∩H0

d = Z(G)0 ∩ Z(Gd). This is equivalent to: ρ exponentiates toa character of H0. By the right hand side of the equality this is independentof the Cartan, and we say G is real-admissible if it holds.

44

This condition is empty unless Z(G) contains a compact torus, and it isweaker than admissibility. For example GL(2,C) is not admissible; GL(2,R)is real-admissible, and U(1, 1) is not.

In any event we can always choose the restriction of λ to Z0(G) to beunitary, so that (7.8)(d) holds.

Example 7.11 Let G = U(1, 1) (see Example 3.4). Then G is not real-admissible, since −I ∈ Z(G)0 and eρ(−I) = −1. Let T be the compactdiagonal Cartan subgroup, and write X∗(T ) = Z2 in the usual coordinates.

Let G be an admissible cover of G. Recall (cf. Example 3.4) X∗(T ) = 12Z⊕Z

or Z ⊕ 12Z.

Suppose χ = (x, y) ∈ X∗(T ) and χ = (a, b) ∈ X∗(T ). Then (6.2)(a) holdsif and only if a− 2x− 1

2= b− 2y + 1

2, in which case

(7.12) µ = (a− 2x− 1

2, a− 2x− 1

2) ∈ z(C)∗.

Since a, 2x ∈ Z we cannot choose µ to be 0.

The transfer factors satisfy the following invariance property with respectto conjugation by G.

Lemma 7.13 Suppose (h, g) ∈ X ′(H, H). Fix x ∈ G and let x = p(p(x)) ∈G. Then

(7.14) ∆eGG(xhx−1, xgx−1) = ∆

eGG(h, g).

Follows easily from Lemma 6.49.From Lemma 6.19 we see:

Lemma 7.15 The group of characters of G acts simply transitively on theset of transfer factors.

More precisely suppose for i = 1, 2 (χis, χis) ∈ S(Hs), and let ∆

eGG(χis, χ

is)

denote the corresponding transfer factors. Then there is a character ψ : G→C× so that for all (h, g) ∈ X ′(G, G)

(7.16) ∆eGG(χ1

s, χ1s)(h, g) = ψ(h)∆

eGG(χ2

s, χ2s)(h, g)

Conversely, suppose that ψ : G → C× is a character and (χ1s, χ

1s) ∈ S(Hs).

Then there exists (χ2s, χ

2s) ∈ S(Hs) such that (7.16) is satisfied.

45

Suppose (h, g) ∈ X(H, H) and eρ(h) is defined. This happens, for exam-ple, if G is admissible, or if Gd is admissible and h ∈ Gd. Then

(7.17)∆1(Φ+, h)ǫr(h,Φ

+)

∆1(Φ+, g)ǫr(g,Φ+)=

∆(h,Φ+)ǫr(h,Φ+)

∆(g,Φ+)ǫr(g,Φ+)eρ(h).

Together with (6.9) this gives simpler formulas for the transfer factor in somesituations. For example (5.20) holds for (h, g) if Gd is acceptable and h ∈ Gd.

Example 7.18 Let G = G = GL(n,R) and let H be the diagonal splitCartan subgroup of G. Choose (χ, χ) as in Example 6.13, and use other

notation as in that Example. Suppose (h, g) ∈ X ′(H, H) and write h =th0 ∈ Γ(H)H0. An easy calculation gives

(7.19) ∆1(Φ+, h)eρ(h0) = |D(h)| 12 .

From this and (6.14) we compute

(7.20) ∆(h, g) =|D(h)| 12|D(g)| 12

τ(h0, g).

This agrees with the transfer factors of [19].

Lemma 7.21 Let H be a Cartan subgroup of G. Suppose (h, g) ∈ X ′(H, H).

Also suppose z ∈ Z(G), z ∈ Z(G) and (z, z) ∈ X(H, H). Then (zh, zg) ∈X ′(H, H) and

(7.22) ∆eGG(zh, zg) = χs(z)/χs(z)∆

eGG(g, h).

Proof. Let Φ+ be a special set of positive roots and choose (χ, χ) ∈S(H,Φ+, χs, χs). Since z and z are central

∆1(Φ+, zh) = ∆1(Φ+, h), ∆1(˜g,Φ+) = ∆1(Φ+, g).

Furthermore

(7.23) Γ(H,Φ+)(zh, zg) = Γ(H,Φ+)(z, z)Γ(H,Φ+)(h, g)

and Γ(H,Φ+)(z, z) = χs(z)/χs(z) by (6.33). Inserting this into the definition

of ∆eGG

gives the result. �

46

8 Some Constants

We need to take care of some constants. Fix an admissible triple (G, G,G).Suppose H is a Cartan subgroup of G, with corresponding Cartan subgroupsH of G and H of G. Let φH be the restriction of φ to H. Define

Definition 8.1

(8.2)c(H) = |Ker(φH)||H/Z(H)|− 1

2 |Z(H)/p−1(φ(H))|−1

= |Ker(φH)||H/Z0(H)|− 12 |Z0(H)/φ(H)|−1

Fix a maximally split Cartan subgroup Hs of G, and define

(8.3) c = c(Hs), C(H) = c(H)/c(Hs).

If it is necessary to specify G we write cG(H), cG = cG(Hs) and CG(H) =cG(H)/cG(Hs).

Let H2 = {h ∈ H | h2 = 1} and H02 = H0 ∩ H2. Note that |H0

2 | =2dim(H∩K). Let Hs, Hf be maximally split and fundamental Cartan sub-groups, respectively.

Proposition 8.4(1) cG(H) = |H0

2 ||H/Z0(H)| 12 .(2) cG(H) = cG(H) |Γ(H)∩C|

|C|.

(3) c(H) and C(H) are integers, and are powers of 2.

(4) c = c(Hs) ≤ c(H) ≤ c(Hf).

(5) 1 ≤ C(H) ≤ C(Hf)

Proof.By definition we have

(8.5)

cG(H) = |Ker(φH)||H/Z0(H)|− 12 |Z0(H)/φ(H)|−1

= |Ker(φH)||H/Z0(H)|− 12 |H/φ(H)|−1|H/Z0(H)|

= |Ker(φH)||H/Z0(H)| 12 |H/φ(H)|−1.

If G = G then Ker(φH) = H2, φ(H) = H0, and

(8.6) cG(H) = |H2||H/Z0(H)| 12 |H/H0|−1.

47

It is easy to see H ≃ H0 ×H/H0, and H/H0 is a two-group, which implies

|H2| = |H02 ||H/H0|. Therefore cG(H) = |H0

2 ||H/Z0(H)| 12 , which is (1).By (1) and (8.5)

(8.7)(a)

cG(H)/cG(H) =|Ker(φH)||Z0(H)/φ(H)|−1

|H2||Z0(H)/H0|−1=

|Ker(φH)||φ(H)/H0||H2|

Recall (Lemma 3.11) φ(H) = (Γ(H) ∩ C)H0. The composition of maps

Hφ

H→ φH(H) → φH(H)/H0 gives an exact sequence

(8.7)(b) 1 → Ker(φH)

Ker(φH) ∩H0 → H

H0 → φH(H)

H0→ 1

Solving the resulting identity for |φH(H)/H0|, inserting in (8.7)(a) andcancelling terms gives

(8.7)(c) cG(H)/cG(H) = |H/H0||Ker(φH) ∩H0|/|H2|.

It is easy to see that Ker(φH) ∩ H0

= H02/C ∩ H0, so the right hand side

equals

(8.7)(d)|H/H0|

|H2/H02 ||C ∩H0|

On the other hand there is an exact sequence

(8.7)(e) 1 → C

C ∩H0→ H

H0→ H

H0 → Γ(H) ∩ C → 1.

The map to Γ(H) ∩ C is given by h → y−1σ(y) where y ∈ H(C) satisfiesp(y) = h. This gives

(8.7)(f)|H/H0||H/H0| =

|Γ(H) ∩ C||C ∩H0||C| .

Using the fact that H/H0 ≃ H2/H02 and plugging this into (8.7)(d) gives (2).

We will see in Lemma 10.8 that |H/Z0(H)| 12 is an integer. Further,H/Z0(H) is a quotient of H/H0, which is a two group, so by (1) cG(H)

48

is an integer and a power of 2. The general case of (3) follows from (1) and(2) if we can show

(8.8) |H02 ||Γ(H) ∩ C|/|C| ∈ Z.

To see this note that C ⊂ Γ(H)H02 . For g ∈ C write g = γh accordingly, and

define ψ(γh) = h(H02 ∩ C). This is a well defined homomorphism from C to

H02/H

02 ∩C. It factors to an inclusion C/C ∩Γ(H) → H0

2/H02 ∩C, and (8.8)

follows from this. This proves (3) for cG(H).It is enough to prove (4); assertion (3) for CG(H) and (5) follow easily.

For this we use Cayley transforms. Suppose Hα, Hβ are as in Section 4. It isenough to show

(8.9) cG(Hα) ≤ cG(Hβ).

First assume G = G. Write Hα = TαAα, Hβ = TβAβ. Then dim(Tβ) =dim(Tα) + 1, and by part (1) of the Lemma it is enough to show

(8.10) |Hα/Z0(Hα)|12 ≤ 2|Hβ/Z0(Hβ)|

12 .

If α is of type II choose t ∈ Hα satisfying α(t) = −1; otherwise let t = 1.By (4.13) write Hα = 〈(Hα ∩Hβ)Bα, t〉. By (4.26)(a)

(8.11)

Hα

Z0(Hα)≃ 〈(Hα ∩Hβ)Bα, t〉

[Z0(Hα) ∩ Z0(Hβ)]Bα

≃ 〈Hα ∩Hβ, t〉[Z0(Hα) ∩ Z0(Hβ)][〈Hα ∩Hβ, t〉 ∩ Bα]

≃ 〈Hα ∩Hβ, t〉Z0(Hα) ∩ Z0(Hβ)

since 〈Hα ∩Hβ, t〉 ∩ Bα ⊂ 〈Hβ, t〉 ∩Bα = 1. Similarly

(8.12)

Hβ

Z0(Hβ)≃ (Hα ∩Hβ)Bβ

[Z0(Hα) ∩ Z0(Hβ)]Bβ

≃ Hα ∩Hβ

[Z0(Hα) ∩ Z0(Hβ)][Hα ∩Hβ ∩Bβ ]

≃ Hα ∩Hβ

〈Z0(Hα) ∩ Z0(Hβ), mα〉

49

We have mα ∈ H0β ⊂ Z0(Hβ). By (4.6)(b) mα ∈ Z0(Hα) if and only if α is

of type I. By (8.11) and (8.12) we see |Hα/Z0(Hα)| = τ |Hβ/Z0(Hβ)| whereτ = 1 (resp. 4) if α is of type I (resp. II). This proves (8.10).

Now supppose G is not equal to G. By (2) of the Lemma

(8.13)cG(Hα)

cG(Hβ)=cG(Hα)

cG(Hβ)

|Γ(Hα) ∩ C||Γ(Hβ) ∩ C|

We need to show the right hand side is ≤ 1.By the preceding argument the first quotient on the right hand side is

equal to 1 if α is of type II for Hα, or 12

otherwise. It is clear that Γ(Hβ) ⊂Γ(Hα) of index 1 or 2, so Γ(Hβ) ∩ C is of index 1 or 2 in Γ(Hα) ∩ C. Weneed to show that if α is of type II for Hα then Γ(Hα) ∩ C = Γ(Hβ) ∩ C.

By the assumption on α we can choose t ∈ Γ(Hα) such that α(t) =−1. Since α(g) = 1 for all g ∈ Γ(Hβ), t 6∈ Γ(Hβ), so Γ(Hα) = 〈Γ(Hβ), t〉.Therefore α(g) = −1 for all g ∈ Γ(Hβ)t, so Γ(Hα) ∩ C = Γ(Hβ) ∩ C.

�

In fact we have shown that

(8.14)cG(Hα)

cG(Hβ)=

{1 α type II for Hα

1 or 12

α type I for Hα

(Only 1/2 occurs in the second case if C = 1). This implies that for anyCartan subgroup H

(8.15) CG(H) ≤ 2r(Hs)−r(H)

where r(∗) denotes split rank.

Example 8.16 If G = GL(n,R) every real root α is of type II, so cG(H) isindependent of H and CG(H) = 1 for all H . See [19].

On the other hand if G = G = U(p, q) then every real root is of type I,and equality holds in (8.15).

9 Lifting: Definition and Basic Properties

Assume that (G, G,G) is an admissible triple. In this section we define lifting

from G to G and derive some of its basic properties. Fix lifting data (χs, χs)

with corresponding transfer factors ∆eGG

(Definition 7.1).

50

Recall (Section 3) O(G, g) is the conjugacy class of g ∈ G, and

(9.1) Ost(G, g) = {xgx−1 | x ∈ G(C)} ∩G

is the stable orbit. We also have O(G, g) and Ost(G, g), and the map (Lemma3.6(3)) φ : Orbst(G) → Orbst(G).

Definition 9.2 Suppose g is a strongly regular semisimple element of G. LetH = p−1(CentG(p(g))). We say g and O(G, g) are relevant if g ∈ Z(H).

The reason for this terminology is:

Lemma 9.3 ([2], Proposition 2.7) Let π be a genuine admissible repre-

sentation of G. Suppose g is a strongly regular semisimple element which isnot relevant. Then Θeπ(g) = 0.

This is elementary, and goes back to GL(2) [9].Here are a few basic properties about orbits.

Lemma 9.4 Suppose g is a semisimple element of G. Then

1. p(O(G, g)) = O(G, p(g))

2. p−1(O(G, p(g))) = O(G, g) ∪O(G,−g)

3. Suppose g is relevant and strongly regular. Then O(G, g) 6= O(G,−g).

Proof. Part (1) and (2) are routine. For (3) suppose not. Then xgx−1 = −gfor some x ∈ G. Let H = CentG(p(g)). Then p(x)p(g)p(x−1) = p(g), so

p(x) ∈ H . But then xgx−1 = g since g ∈ Z(H) and x ∈ H. �

Lemma 9.5 Suppose Ost ∈ Orbst(G) and write φ(Ost) = Ost(G, g) for some

g ∈ G. Assume g is strongly regular. Then there is a unique h ∈ Ost

suchthat φ(h) = g. Furthermore h ∈ H and is also strongly regular.

Proof. By definition there are h′ ∈ Ostand g′ ∈ Ost(G, g) such that φ(h′) =

g′. Let x ∈ G(C) such that g′ = xgx−1. Since g, g′ ∈ G we have σ(x)−1x ∈CentG(C)(g) = H(C). Let y = p(x) ∈ G(C) and h = y−1h′y. Since g =φ(h) = s(h)2, s(h) ∈ CentG(C)(g) = H(C). But σ(y)−1y ∈ H(C) so σ(h) = h

and h ∈ H ∩ Ostwith φ(h) = g.

51

Suppose r ∈ CentG(C)(h), and choose a preimage s of r in G(C). Then

g = φ(h) = φ(rhr−1) = sφ(h)s−1 = sgs−1 so that s ∈ H(C). Thus r ∈ H(C),so that h is also strongly regular.

Suppose that h1 ∈ Ostwith φ(h1) = g. Then there is u ∈ G(C) with

uhu−1 = h1. Choose a preimage v of u in G(C). Then g = φ(h1) =φ(uhu−1) = vφ(h)v−1 = vgv−1. Thus v ∈ H(C). Now u ∈ H(C) so h1 = h.�

Remark 9.6 Note that Lemma 9.5 implies the following. Let Ost ∈ Orbst(G)

and Ost ∈ Orbst(G) with φ(Ost) = Ost. If Ost is strongly regular and

semisimple, then so is Ost, and the restriction of φ to Ost

gives a bijection

between Ostand Ost. Further, for any g ∈ Ost ∩H ,

(9.7) {Ost |φ(Ost) = Ost} = {Orbst(G, h) | h ∈ H, φ(h) = g}.

Definition 9.8 Suppose O ∈ Orb(G) and Ost ∈ Orbst(G) are strongly regu-

lar and semisimple. Let p(O)st be the stable orbit for G containing p(O).

Define ∆eGG(Ost

, O) = 0 unless φ(Ost

) = p(O)st.

Suppose φ(Ost

) = p(O)st. Choose g ∈ O. By Lemma 9.5 there is a unique

h ∈ Ostwith φ(h) = p(g). Define

(9.9) ∆eGG(Ost

, O) = ∆eGG(h, g).

By Lemma 7.13 this is independent of the choice of g.

Recall a virtual character π of G is said to be stable if the function θπrepresenting its character is constant on the stable orbit Ost(G, g) for allstrongly regular semisimple elements g.

Definition 9.10 Suppose Θ is a stable virtual character of G. Then Θ(Ost

)

is defined for any strongly regular semisimple orbit Ost ∈ Orbst(G). Suppose

O ∈ Orb(G) is strongly regular and semisimple. Recall c = cG is given byDefinition 8.1.

Define

(9.11)(a) LifteGG(Θ)(O) = c−1

∑

{Ost

∈Orbst(G)}

∆eGG(Ost

, O)Θ(Ost).

52

This is a finite sum, over

(9.11)(b) {Ost |φ(Ost

) = p(O)st}.

Thus LifteGGΘ is a genuine class function defined on the set of strongly

regular semisimple orbits in G. The following lemma follows easily from (9.7)and the fact that φ restricted to H is a homomorphism. Recall X(H, g) isgiven by Definition 5.2.

Lemma 9.12 Suppose g is a strongly regular semisimple element. Let H =CentG(p(g)) and H = p(H). Then

(9.13) LifteGG(Θ)(g) = c−1

∑

{h∈X(H,eg)}

∆eGG(h, g)Θ(h).

In particular LifteGG(Θ)(g) = 0 unless p(g) is in the image of φ. Assume this

holds, and choose h satisfying φ(h) = p(g). Then

(9.14) LifteGG(Θ)(g) = c−1

∑

{t∈H |φ(t)=1}

∆eGG(th, g)Θ(th).

Remark 9.15 Formula (9.13) can be used to extend LifteGG(Θ) to a genuine

class function on all regular (not just strongly regular) semisimple elements.

We derive some elementary properties of lifting. Recall the definitions of(5.1).

Lemma 9.16 LifteGG(Θ) is supported on Z(G)G0

d.

Proof. Let H be a Cartan subgroup of G. Then H = Z(G)Γ(H)H0

d, so that

φ(H) ⊂ Z0(G)H0d . Thus p−1φ(H) ⊂ Z(G)G0

d. �

Define Z1(G) = {z ∈ Z(G) : φ(z) = 1}.

Lemma 9.17 Let Θ be a stable character of G with central character ζ.

Then LifteGG(Θ) ≡ 0 unless ζ(z) = χs(z) for all z ∈ Z1(G).

Proof. Fix a Cartan subgroup H , g ∈ H ′, and z ∈ Z1(G). Then

{h ∈ H |φ(h) = p(g)} = {zh | h ∈ H, φ(h) = p(g)}.

53

Further, for all h ∈ H such that φ(h) = p(g),

∆eGG(g, zh) = χs(z

−1)∆eGG(g, h)

by Lemma 7.21 applied to h, g, z, and z = 1. We have

LifteGG(Θ)(g) = c−1

∑

X(H,eg)

∆eGG(g, zh)Θ(zh) = c−1ζ(z)χs(z

−1)LifteGG(Θ)(g).

�

Fix a character ζ of Z(G) satisfying ζ(z) = χs(z) for all z ∈ Z1(G). For

z ∈ S = p−1φ(Z(G)) define

(9.18) ζ(z) = χs(z)(ζχ−1s )(z)

where z ∈ Z(G), φ(z) = p(z). This is independent of the choice of z, and is

a genuine character of S.

Lemma 9.19 Let Θ be a stable character of G with central character ζ suchthat ζ(z) = χs(z) for all z ∈ Z1(G). Then for all g ∈ G, z ∈ S,

LifteGG(Θ)(zg) = ζ(z)Lift

eGG(Θ)(g).

Proof. Fix a Cartan subgroup H of G, g ∈ H ′ and z ∈ S. Choose z ∈ Z(G)such that φ(z) = p(z). Then

{h ∈ H | (h, zg) ∈ X(H, H)} = {zh | h ∈ H, (h, g) ∈ X(H, H)}.

Thus by Lemma 7.21,

LifteGG(Θ)(zg) = c−1

∑

X(H,eg)

∆eGG(zg, zh)(Θ(zh)

= (χs(z)/χs(z))ζ(z)LifteGG(Θ)(g) = ζ(z)Lift

eGGΘ)(g).

�

54

10 Lifting for Tori

We give some details of lifting for tori. This illustrates some of the basicprinciples, and plays an important role in lifting for general reductive groups.

Let G(C) be an algebraic torus, with real points G. Let p : G → G beany two-fold cover; any such cover is necessarily admissible (Definition 3.3).

Let C be a subgroup of p(Z(G)) consisting of elements of order 2, and let

G = G(C)/C, with real points G. Then (G, G,G) is an admissible triple.Since G is a torus, G = H = Hs is a maximally split Cartan subgroup.

Let Xg(Z(G)) denote the genuine characters of Z(G) and let X(G) denotethe characters of G. Since gd = 0 condition (6.2) is empty, so (χ, χ) ⊂ S(G)

for all χ ∈ Xg(Z(G)), χ ∈ X(G).

The transfer factor ∆eGG

of Definition 7.1 is then given by

∆eGG(h, g) = χ(g)χ(h)−1 ((h, g) ∈ X(G, G)).

Let

(10.1) S = {g ∈ G | p(g) ∈ φ(G)} = {g |X(G, g) 6= ∅}.

If ψ ∈ X(G) then

(10.2) LifteGG(ψ)(g) = c−1χ(g)

∑

h∈X(G,eg)

χ−1(h)ψ(h),

with c ∈ Z given by Definition 8.1. In particular LifteGG(ψ)(g) = 0 for g 6∈ S.

Suppose g ∈ S and fix h ∈ X(G, g). Then

(10.3) LifteGGψ(g) = c−1χ(g)χ−1(h)ψ(h)

∑

{t∈G |φ(t)=1}

χ−1(t)ψ(t).

Thus LifteGGψ = 0 unless χ(t) = ψ(t) for all t ∈ Ker(φ). Assume this holds.

For g ∈ S choose h ∈ X(G, g) and define

(10.4) ψ0(g) = χ(g)ψ(h)χ(h)−1.

This is independent of the choice of h ∈ X(G, g), so ψ0 is a well defined

genuine character of S, and it follows immediately that

55

(10.5) LifteGG(ψ)(g) =

{0 g 6∈ S

c−1|Ker(φ)|ψ0(g) g ∈ S.

It follows easily from the induced character formula that

(10.6) LifteGG(ψ) = c−1|Ker(φ)||G/S|−1 Ind

eGeS(ψ0).

By induction by stages we have

(10.7) IndeGeS(ψ0) = Ind

eGZ( eG)

IndZ( eG)eS

(ψ0)

In order to identify this as a sum of characters of G, we use the followingelementary result.

Lemma 10.8 ([1], Proposition 2.2) Let ψ ∈ Xg(Z(G)). Then there is a

unique irreducible genuine representation τ = τ (ψ) of G such that τ |Z( eG) is a

multiple of ψ. The map ψ → τ (ψ) is a bijection between Xg(Z(G)) and the

set of equivalence classes of irreducible genuine representations of G, and

IndeGZ( eG)

ψ = |G/Z(G)| 12 τ(ψ).

The dimension of τ(ψ) is |G/Z(G)| 12 ; in particular this is an integer.

Let

(10.9) Xg(Z(G), ψ0) = {ψ ∈ Xg(Z(G)) | ψ|eS = ψ0}.

By Frobenius reciprocity we have

(10.10) LifteGG(ψ) = c−1|Ker(φ)||G/S|−1|G/Z(G)| 12

∑

eψ∈Xg(Z( eG), eψ0)

Θeτ( eψ).

By (8.2) the term in front of the sum is equal to 1 (of course we defined cprecisely to make this hold).

We summarize these results. As defined, LifteGG(ψ) is a class function on

G, which by (10.10)) the character of a representation. We identify LifteGG(ψ)

with this representation.

56

Proposition 10.11 Fix χ ∈ Xg(Z(G)) and χ ∈ X(G), and use them to

define LifteGG. Fix ψ ∈ X(G). Then Lift

eGG(ψ) 6= 0 if and only if χ(t) = ψ(t) for

all t ∈ Ker(φ). Assume this holds. Define ψ0 as in (10.4) and Xg(Z(G), ψ0)as in (10.9). Then

(10.12) LifteGG(ψ) =

∑

eψ∈Xg(Z( eG), eψ0)

τ (ψ).

This is an identity of genuine representations of G; the right hand side is thedirect sum of |p(Z(G))/φ(G)| irreducible representations. The differentialssatisfy

(10.13) dψ =1

2(dψ − dµ)

where dµ = dχ− 2dχ ∈ g(C)∗ as in (6.36)(c).

In particular assume C is chosen so that φ(G) = p(Z(G)). Then Xg(Z(G), ψ0)

consists of the single character ψ0, and

(10.14) LifteGG(ψ) = τ(ψ0).

If it is necessary to indicate the dependence of LifteGG on χ and χ we will

write

(10.15) LifteGG(χ, χ, ψ) = Lift

eGG(ψ).

Corollary 10.16 Fix (χ, χ) as above, and let ψ be a genuine character of

Z(G). Then there is a unique character ψ of G such that τ(ψ) occurs in

LifteGG(χ, χ, ψ). It is given by

(10.17) ψ(h) = χ(h)(ψχ−1)(φ(h)) (h ∈ G).

Example 10.18 Let G(C) be an algebraic torus with connected real points

G and let G be a two-fold cover. Then G = Z(G) is also abelian. We may

as well take G = G. Let χ be any genuine character of G, and let χ = χ2.

Suppose ψ ∈ X(G). Using Proposition 10.11 LifteGG(ψ) 6= 0 if and only if

(10.4) gives a well-defined genuine character ψ of G. It is easy to see that in

this case ψ is the unique genuine character with ψ = ψ2.Therefore, with this choice of (χ, χ), Lift

eGG(ψ) 6= 0 if and only if ψ has a

genuine square root ψ on G, in which case LifteGG(ψ) = ψ.

57

11 Example: Minimal Principal Series of Split

Groups

Now assume (G, G,G) is an admissible triple where G is split. Let Hs be asplit Cartan subgroup of G, let Φ denote the roots of Hs, and let W = W (Φ).Fix a set of positive roots Φ+ and choose (χs, χs) ∈ S(G,Hs,Φ

+) (cf. 6.4).

Let ρ = ρ(Φ+) and define ∆eGG

as in Definition 7.1 accordingly.

Corresponding to each character ψ of Hs is a principal series representa-tion π(ψ) of G. Its character Θπ(ψ) is supported on the conjugates of Hs andsatisfies

(11.1) Θπ(ψ)(h) = |DG(h)|− 12

∑

w∈W

ψ(wh) (h ∈ H′

s).

We wish to compute LifteGG(π(ψ)). Using (9.13) it is clear that Lift

eGG(Θπ(ψ))(g) =

0 unless g is conjugate to an element of Hs.Suppose g ∈ H ′

s. Then by (9.13)

(11.2) LifteGG(Θπ(ψ))(g) = c−1

∑

h∈X(Hs,eg)

∆eGG(h, g)||DG(h)|− 1

2

∑

w∈W

ψ(wh).

Suppose h ∈ X(Hs, g). From the definitions we have

(11.3) ∆eGG(h, g)|DG(h)|− 1

2 = |D eG(g)|− 12 |eρ(h)|χs(g)χ−1

s (h).

If w ∈W then wh ∈ X(Hs, wg) and by Lemma 7.13 we have

(11.4) ∆eGG(h, g)|DG(h)|− 1

2 = ∆eGG(wh,wg)|DG(wh)|− 1

2 .

Therefore LifteGG(Θπ(ψ))(g) =

(11.5) c−1∑

w∈W

|D eG(wg)|− 12 χs(wg)

∑

h∈X(Hs,eg)

|eρ(wh)|χ−1s (wh)ψ(wh).

Since χs ∈ Xg(Z(Hs)) and χs|e−ρ| ∈ X(Hs), we can use the pair (χs, χs|e−ρ|)to define Lift

eHs

Hsas in §10. By Proposition 10.11 and a short calculation, we

can write

(11.6) LifteGG(Θπ(ψ)(g) = |D eG(g)|− 1

2

∑

w∈W

LifteHs

Hs(ψ))(wg).

58

By Proposition 10.11 the lift is non-zero if and only if

(11.7) ψ(h) = |e−ρ(h)|χs(h) (h ∈ Hs, φ(h) = 1).

If h = mα then by (6.2)(b) the right hand side is trivial (ζcx = 1 since there

are no complex roots). Therefore LifteGG

= 0 independent of the choice oftransfer factors unless ψ(mα) = 1 for all α ∈ Φ.

Assume (11.7) holds. We apply Proposition 10.11 to the summands of

(11.6). Let S = p−1(φ(Hs)) ⊂ Z(Hs) and let n = |Z(Hs)/S|12 . Define a

character of S:

(11.8) ψ0(g) = χs(g)ψ(h)χs(h)−1|eρ(h)|

where φ(h) = p(g). Write

(11.9) {ψ ∈ Xg(Z(Hs)) | ψ|eS = ψ0} = {ψ1, . . . , ψn}.

By Proposition 10.11

(11.10) LifteHs

Hs(ψ) =

n∑

i=1

π(τ (ψi))

and plugging this into (11.6) gives

(11.11) LifteGG(Θπ(ψ))(g) = |D eG(g)|− 1

2

n∑

i=1

∑

w∈W

Θeτ( eψi)

(wg).

It is straightforward to identify the right hand side of (11.11) with a sum ofprincipal series characters.

Suppose τ is an irreducible genuine representation of Hs. Associated toτ is a genuine principal series representation π(τ) of G; see Section 16 for

details. Then Θπ(eτ ) is supported on the conjugates of Z(Hs), and for g ∈ H ′s

we have

(11.12) Θπ(eτ)(g) = |D eG(g)|− 12

∑

w∈W

Tr(τ (wg)).

Write τ = τ (ψ) for ψ a genuine character of Z(Hs) as in Lemma 10.8, and

let π(ψ) = π(τ (ψ)). Comparing (11.11) and (11.12) we obtain the followingresult. Let µ = µ(χs, χs) = dχs − 2dχs − ρ ∈ z(C)∗ (Definition 6.35).

59

Proposition 11.13 Fix (χs, χs) ∈ S(Hs) for (G, G,G), and use this to de-

fine LifteGG. Define Lift

eHs

Hsusing (χs, χs|e−ρ|) ∈ S(Hs) for (Hs, Hs, Hs). Sup-

pose ψ ∈ X(Hs). Then LifteGG(π(ψ)) = 0 unless (11.7) holds, so assume this

is the case. Define µ, S, n = |Z(Hs)/S| and {ψ1, . . . , ψn} as above. Then

(11.14) LifteGG(π(ψ)) =

n∑

i=1

π(ψi).

Each term in the sum has infinitesimal character 12(dψ − µ).

Conversely, for ψ ∈ Xg(Z(Hs)) define

(11.15) ψ(h) = |e−ρ(h)|χs(h)(ψχ−1s )(φ(h)) (h ∈ H).

Then Θπ( eψ) occurs in LifteGG(π(ψ)).

Remark 11.16 Note that H0s ⊂ S ⊂ Z(Hs) = Z(G)H0

s . This implies

dψi = 12(dψ−µ) (independent of i) and that the restrictions of the ψi to Z(G)

are distinct. Therefore the principal series representations π(ψi) occuring in(11.14) have distinct central characters, and are a fortiori not isomorphic.

Example 11.17 Suppose G(C) is simple and simply connected. In this

case lifting is canonical (cf. Section 5) and µ = 0. Also Z(G) = Z(G) and

|Z(Hs)/H0s | = |Z(G)|. In fact |Z(G)| = 4 (in type D2n), 2 (in type A2n+1,

D2n+1 or E7) or 1 otherwise. See [1, Table I].

Suppose G = G. Then S = H0s and n = |Z(G)|. By (11.7) Lift

eGG(π(ψ)) 6=

0 if and only if π(ψ) is spherical. Thus the lift of the spherical principal seriesrepresentation with infinitesimal character λ is the sum of the n genuineprincipal series representations of G with infinitesimal character λ/2.

On the other hand suppose C = Z(G); this is allowed by Lemma 3.15.

Then S = Z(H), and the lift consists of a single principal series.

Example 11.18 We specialize the preceding example. Suppose G = G =SL(2,R). Write Hs ≃ R× and suppose ψν(x) = |x|ν . Then Hs ≃ R× ∪ iR×.

Suppose ψ±ν (x) = |x|ν and ψ±

ν (i) = ±i. Then

(11.19) LifteGG(π(ψν)) = π(ψ+

ν/2) + π(ψ−ν/2).

60

Note that Z(G) = {±1,±i} and ψ±ν have different central characters. Also

note that the character of π(ψ+ν ) + π(ψ−

ν ) vanishes on iR×. In this caseφ(Hs) = H0

s .Now take G = PSL(2,R) ≃ SO(2, 1). Then (see Example 1.11) φ :

Hs → Hs is an isomorphism. By (11.7) the lift of any principal series of Gis non-zero. With the obvious notation we have

(11.20) LifteGG(π(ψ±

ν )) = π(ψ±ν/2).

12 Example: Discrete Series on the Compact

Cartan

Suppose G = G is connected, semisimple, acceptable, and equal rank. Weassume also that G is connected. Let H be a compact Cartan subgroup of G.We compute the character formula for the lift of discrete series representationrestricted to H .

The group H is connected. Suppose λ ∈ h(C)∗ satisfies 〈λ, α∨〉 ∈ Z6=0 forall roots α. Associated to λ is a stable sum of discrete series, πst(λ) of G,whose character Θπst(λ) satisfies

(12.1) Θπst(λ)(h) = (−1)q∆(Φ+, h)−1∑

w∈W

ǫ(w)ewλ(h) (h ∈ H ′).

Here Φ+ = Φ+(λ) = {α | 〈λ, α∨〉 > 0} and q = 12dim(G/K).

Suppose (h, g) ∈ X(H, H). By Definition 5.17 (see (5.20)) the transferfactors are canonical, and given by

(12.2) ∆(h, g) =∆(h,Φ+)

∆(g,Φ+)τ(h, g).

By (12.1) and (9.13) we have:(12.3)

LifteGG(Θπst(λ))(g) = c−1(−1)q∆(g,Φ+)−1

∑

W

ǫ(w)∑

h∈X(H,eg)

ewλ(h)τ(h, g).

Define LifteHH using lifting data (χ, χ2) as in Example 10.18. Then the

inner sum is equal to c(H)LifteHH(ewλ)(g). By Definition 8.1 C(H) = c(H)/c,

61

so

(12.4) LifteGG(Θπst(λ))(g) = C(H)(−1)q∆(g,Φ+)−1

∑

W

ǫ(w)LifteHH(ewλ)(g).

By Example 10.18 LifteHH(ewλ) 6= 0 if and only if ewλ/2 is a genuine charac-

ter of H . Let W# be the set of w ∈W for which this holds. Assume W# 6= ∅.By replacing λ with wλ we may assume 1 ∈W#. Then

(12.5) W# = {w ∈ W |wλ2− λ

2∈ X∗(H)}

where X∗(H) is the character lattice of H . From Lemma 3.2 it is clear that

(12.6) W (G,H) = {w ∈W |wλ2− λ

2∈ R} ⊂ W#.

where R is the root lattice. A short calculation gives

(12.7) LifteGG(Θπst(λ))(g) =

C(H)∑

w∈W (G,H)\W#

(−1)q∆(g,Φ+(wλ))−1∑

v∈W (G,H)

ǫ(v)evwλ/2(g).

For each w ∈ W# the summand is the formula for the character of

π(wλ/2), the genuine discrete series representation of G with Harish-Chandraparameter wλ/2. Therefore

(12.8) LifteGG(Θπst(λ))(g) = C(H)

∑

w∈W (G,H)\W#

Θπ(wλ/2)(g) (g ∈ H ′).

Since formulas for discrete series characters are more complicated on non-

compact Cartan subgroups, we will not directly compute LifteGG(Θπst(λ))(g) for

g in an arbitrary Cartan subgroup. In Section 14 we will use results of Hiraiand Harish-Chandra to show that the lift of a stable, supertempered eigendis-tribution on G is a supertempered eigendistribution on G. Thus both sidesof (12.8) are supertempered eigendistributions. Now a theorem of Harish-

Chandra that says if two supertempered eigendistributions on G agree onthe compact Cartan subgroup, they are equal. This allows us to concludethat (12.8) is valid for all g ∈ G′. See Section 18.

62

Remark 12.9 If w ∈ W (G,H)\W# then w λ2− λ

2∈ X∗(H)/R ≃ Z(G(C)).

Since G is of equal rank Z(G(C)) = Z(G). Therefore π(wλ/2) and π(λ/2)have distinct central characters, and the terms in (12.8) are a fortiori notisomorphic. Compare Remark 11.16.

Example 12.10 Let G = Spin(2p, 2q), and write λ = (a1, . . . , ap; b1, . . . , bq)in the usual coordinates. Then the lift is non-zero if and only if all ai are even,and all bi are odd, or vice-versa. If p 6= q then the sum is a singleton, and

LifteGG(π(λ)) = C(H)π(λ/2). If p = q then the sum consists of two elements,

λ/2 and λ′/2 = 12(b1, . . . , bp; a1, . . . , ap).

Example 12.11 Suppose G is compact, so G ≃ G×Z/2Z (which is admis-sible, but not connected). Then π(λ) is the finite dimensional representation

with infinitesimal character λ. The calculation of LifteGG(π(λ)) is the same

as above, except that LifteHH(ewλ) 6= 0 if and only if ewλ/2 is a character of

H , in which the genuine square root of ewλ is ewλ/2 ⊗ sgn. In this caseW = W (G,H) = W# so the sum on the right hand side of (12.8) has oneelement. Thus

(12.12) LifteGG(π(λ)) = π(λ/2) ⊗ sgn

if λ/2 ∈ X∗(H), or is 0 otherwise.

Also see Example 18.26.

13 Invariant Eigendistributions

In this section we review results of Hirai and Harish-Chandra on invarianteigendistributions. In the next section we apply these results to study liftingfrom G to G.

Througout this section we assume G is a group in Harish-Chandra’s class(see the Appendix). This holds for the groups (G, G,G) in an admissibletriple (see the beginning of Section (14)). Let H be a Cartan subgroup ofG. Recall H ′ is the set of regular elements of H and suppose F : H ′ → C isdifferentiable. Corresponding to X ∈ h is the differential operator

(13.1) DXF (h) =d

dt

∣∣∣∣t=0

F (hexp(tX)) (h ∈ H ′).

63

Fix a set of positive roots Φ+, with corresponding ρ = ρ(Φ+). Then we alsodefine

(13.2) DρXF (h) =

d

dt

∣∣∣∣t=0

e<ρ,tX>F (hexp(tX)) (h ∈ H ′).

The map DX → DρX =< ρ,X > +DX can be extended to an automorphism

D → Dρ of S(h(C)).Let ν be a character of the center Z of the universal enveloping algebra

of g(C). Using the Harish-Chandra homomorphism we identify ν with acharacter of I(h(C)) = S(h(C))W .

We say F : H ′ → C satisfies condition (C1, Φ+, ν ) if F is real analyticon H ′ and

(C1,Φ+, ν) DρF (h) = ν(D)F (h) (h ∈ H ′, D ∈ I(h(C))).

Let H ′(R) = {h ∈ H | eα(h) 6= 1, α ∈ Φr}. We say F satisfies condition (C2)if

(C2) F extends to a real analytic function on H ′(R).

Let G′ be the set of regular semisimple elements of G, and suppose Θ isa class function on G′. With ∆1(Φ+, h) defined as in (5.1) let

(13.3) Ψ(H,Φ+, h) = ∆1(Φ+, h)Θ(h) (h ∈ H ′).

Assume Ψ(H,Φ+) satisfies (C1, Φ+, ν ) and (C2).Let α be a simple root of Φ+

r . Let J be the corresponding Cayley trans-form of H as in Section 4. Choose c ∈ Aut(g(C)) satisfing (4.14)(a) and(b), and let β = c∗(α) be the imaginary root of J corresponding to α. LetΦ+J = c∗Φ+ and ρJ = ρ(Φ+

J ). Let H(α) be the set of h ∈ H such thateα(h) = 1, but eγ(h) 6= 1 for all γ 6= ±α.

Fix h ∈ H(α). Then hexp(tα∨) ∈ H ′ for 0 6= t sufficiently small, andthere are well-defined one-sided limits

(13.4) Dρ,±α∨ Ψ(H,Φ+, h) = lim

t→0±Dρα∨Ψ(H,Φ+, hexp(tα∨)).

Note that iβ∨ ∈ j and h ∈ J ′(R). Thus DρJ

β∨ = −iDρJ

iβ∨ is defined as in (13.2),

and DρJ

β∨Ψ(J,Φ+J , h) is defined. We say Ψ(H,Φ+) satisfies condition (C3) if

64

(C3) [Dρ,+α∨ −Dρ,−

α∨ ]Ψ(H,Φ+, h) = 2DρJ

β∨Ψ(J,Φ+J , h) (h ∈ H(α)).

The following theorem was proved by Hirai [15], [17] assuming two extraconditions (see the Appendix) which may fail to hold in our situation. Inthe Appendix we extend Hirai’s theorem to a class of reductive groups thatincludes Harish-Chandra’s class.

Theorem 13.5 (cf. Theorem 20.23) Let Θ be a class function on G′ andlet ν be a character of Z. Then Θ is an invariant eigendistribution withinfinitesimal character ν if and only for every Cartan subgroup H of G andevery real root α of H, there is a choice of positive roots Φ+ such that α issimple for Φ+

r and the function Ψ(H,Φ+) satisfies conditions (C1, Φ+, ν),(C2) , and (C3).

Moreover, suppose Θ is an invariant eigendistribution. Then Ψ(H,Φ+)satisfies (C1, Φ+, ν) and (C2) for every choice of Φ+, and satisfies (C3) forevery choice of Φ+ such that α is simple for Φ+

r .

We now review some results of Harish-Chandra. Write G = Kexp(p)where K is a maximal compact subgroup and g = k ⊕ p is the Cartan de-composition. Let ‖ · ‖G be a Euclidean norm on p and define

(13.6) τG(kexpX) = ‖X‖, k ∈ K,X ∈ p.

Let Θ be an invariant eigendistribution on G. Then by §12 of [11], Θ istempered if and only if for each Cartan subgroup H of G there are numbersC, r ≥ 0, so that

(13.7) |D(h)| 12 |Θ(h)| ≤ C(1 + τG(h))r, h ∈ H ′,

(see 5.1 for the definition of |D(h)| 12 ).Let z be the center of g. As in [11] we can decompose G = 0GZp where

0G has Lie algebra [g, g] ⊕ (z ∩ k) and Zp = exp(z ∩ p). Let Θ be an in-variant eigendistribution on G. We say Θ is relatively tempered (relativelysupertempered) on G if its restriction to 0G is tempered (supertempered) on0G. By §4 of [12], Θ is supertempered on 0G if and only if for every Cartansubgroup 0H of 0G and every r ≥ 0,

(13.8) supg∈0H′

|D(h)| 12 |Θ(h)|(1 + τG(h))r <∞.

65

We will also use the following result from §4 of [12].

Theorem 13.9 [12] Assume that G has a relatively compact Cartan sub-group B and let Θ be a relatively supertempered invariant eigendistributionsuch that Θ(b) = 0, b ∈ B′. Then Θ = 0.

14 Lifting of Invariant Eigendistributions

We now return to the context of lifting. Suppose (G, G,G) is an admissible

triple. We note that each of the groups G, G, and G are in Harish-Chandra’sclass (see the Appendix). For G and G this is immediate. For G note

that Cent eG(G0) = Z(G) by (4.3). Therefore Ad(G) ≃ Ad(G), so G alsosatisfies Condition A of the Appendix, and the remaining conditions areeasy. Therefore the results of Section 13 apply.

Fix a maximally split Cartan subgroup Hs of G and lifting data (χs, χs) ∈S(Hs) as in Definition 7.1. Recall µ ∈ z(C)∗ is given by (6.36)(c). Let Θ be a

stable invariant eigendistribution on G. Define Θ = LifteGGΘ as in Definition

9.10; Θ is a genuine class function on G′.

Theorem 14.1 Let Θ be a stable invariant eigendistribution on G with in-

finitesimal character ν. Then LifteGGΘ is an invariant eigendistribution on G

with infinitesimal character (ν − µ)/2.

We apply Theorem 13.5. We need to show that conditions (C1-3) hold.Let H be a Cartan subgroup of G, and let Φ+ be a choice of positive

roots. Define

(14.2)(a) Ψ(H,Φ+, h) = ∆1(Φ+, h)Θ(h) (h ∈ H′),

(14.2)(b) Ψ(H,Φ+, g) = ∆1(Φ+, g)Θ(g) (g ∈ H ′).

The second part of Theorem 13.5 says:

Lemma 14.3 Ψ(H,Φ+) satisfies conditions (C1, Φ+, ν) and (C2) for everychoice of Φ+. Let α ∈ Φ+

r . Then Ψ(H,Φ+) satisfies condition (C3) for anychoice of Φ+ such that α is a simple root for Φ+

r .

66

We now verify conditions (C1)-(C3) for Ψ(H,Φ+). We begin with (C1).Note that µ defines a character of I(h(C)) = S(h(C))W and also by theHarish-Chandra homomorphism a character of Z. Assume Φ+ is a special setof positive roots.

Proposition 14.4 For all D ∈ I(h(C)) and g ∈ H ′,

DρΨ(H,Φ+, g) = ((ν − µ)/2)(D)Ψ(H,Φ+, g).

In other words Ψ(H,Φ+) satisfies condition (C1,Φ+,(ν − µ)/2).

Since H,H and Φ+ are fixed we write Ψ(g) = Ψ(H,Φ+, g) and Ψ(h) =

Ψ(H,Φ+, h). Let Γ = Γ(H,Φ+) (Definition 6.35). Thus for (h, g) ∈ X(H, H),Γ(h, g) = χ(g)χ−1(h) where (χ, χ) ∈ S(H,Φ+, χs, χs).

Lemma 14.5 For all g ∈ H,X ∈ h such that gexpX ∈ H ′,

Ψ(gexpX) = c−1∑

h∈X(H,eg)

Γ(h, g)e−<ρ+µ,X/2>Ψ(hexp(X/2)).

Proof. Let g ∈ H ′. Recall the definition of X(H, g) (5.3)(c). From thedefinitions it is clear that

Ψ(g) = c−1∑

h∈X(H,eg)

Γ(h, g)Ψ(h).

Fix g ∈ H,X ∈ h such that gexpX ∈ H ′. Then φ(exp(X/2)) = expX so

X(H, gexpX) = {hexp(X/2) : h ∈ X(H, g)}.

Thus

Ψ(gexpX) =∑

h∈X(H,eg)

Γ(h, g)Γ(exp(X/2), expX)Ψ(hexp(X/2).

ButΓ(exp(X/2), expX) = e<2deχ−dχ,X/2> = e−<ρ+µ,X/2>

by (6.36)(c). �

Proof of Proposition 14.4.

67

It follows from Lemmas 14.3 and 14.5 that Ψ is real analytic on H ′.Suppose X ∈ h and t ∈ R. Using Lemma 14.5 we have

e<ρ,tX>Ψ(gexptX) = c−1∑

h∈X(H,eg)

Γ(h, g)e<ρ−µ,tX/2>Ψ(hexp(tX/2)).

ThusDρXΨ(g) = c−1

∑

h∈X(H,eg)

Γ(h, g)(DρX/2− < µ,X/2 >)Ψ(h).

First assume X ∈ z. Then DX/2 ∈ I(h(C)), so DρX/2Ψ(h) = ν(X/2)Ψ(h)

by Lemma 14.3. Thus

DρXΨ(g) = c−1

∑

h∈X(H,eg)

Γ(h, g) < ν−µ,X/2 > Ψ(h) = ((ν −µ)/2)(DX)Ψ(g).

By induction we see that

(14.6) DρΨ(g) = ((ν − µ)/2)(D)Ψ(g) for all D ∈ S(z(C)).

Now suppose X ∈ hd. Then < µ,X >= 0, so

(14.7) DρXΨ(g) =

1

2c

∑

h∈X(H,eg)

Γ(h, g)DρXΨ(h).

Now suppose Y ∈ hd, replace g with gexp(tY ), and multiply both sidesby e〈ρ,tY 〉. The right hand side is

(14.8) e〈ρ,tY 〉 1

2c

∑

h∈X(H,eg)

Γ(hexp(tY/2), gexp(tY ))DρXΨ(hexp(tY/2)).

As in the proof of Lemma 14.5 we can write

Γ(hexp(tY/2), gexp(tY )) = Γ(h, g)e〈−ρ,tY/2〉

and we see(14.9)

e〈ρ,tY 〉DρXΨ(gexp(tY )) =

1

2c

∑

h∈X(H,eg)

Γ(h, g)e〈ρ,tY/2〉DρXΨ(hexp(tY/2)).

68

Taking the derivative and evaluating at t = 0 we see

(14.10) DρYD

ρXΨ(g) =

1

4c

∑

h∈X(H,eg)

Γ(h, g)DρYD

ρXΨ(h).

By induction we conclude that for D ∈ I(hd(C)), homogeneous of degree k,we have

(14.11)

DρΨ(g) =1

2kc

∑

h∈X(H,eg)

Γ(h, g)DρΨ(h)

=1

2kcν(D)

∑

h∈X(H,eg)

Γ(h, g)Ψ(h)

= (ν/2)(D)Ψ(g)

= ((ν − µ)/2)(D)Ψ(g).

The last equality comes from the fact that µ(D) = 0.We have I(h(C)) = I(hd(C))S(z(C)), and the result follows from (14.11)

and (14.6). �

We now consider condition (C2). We continue to write Ψ(g) = Ψ(H,Φ+, g)and Ψ(h) = Ψ(H,Φ+, h). Recall

(14.12) H ′(R) = {g ∈ H | eα(g) 6= 1, α ∈ Φr}.

Proposition 14.13 Ψ extends to a real analytic function on H ′(R), i.e. Ψsatisfies condition (C2).

Proof. Since the restriction of Θ to H is supported on Z(G)H0, Ψ is also

supported on Z(G)H0. Fix t ∈ Z(G)T 0. If Φ+r is a set of positive roots of

Φr let

(14.14) C(Φ+r ) = {X ∈ a |α(X) > 0 for all α ∈ Φ+

r }.

Then the connected components of tH0∩H ′(R) are of the form tT 0exp(C(Φ+r )).

Fix h ∈ X(H, t). Note that for all α ∈ Φr, e2α(h) = eα(t) = 1, so

that eα(h) = ±1. Let Φr(h) = {α ∈ Φr | eα(h) = 1}. Then the connected

components of hH0∩H ′

(R) have the form hT0exp(C(Φ+

r (h))) with C(Φ+r (h))

defined as in (14.14).

69

Given Φ+r , let Φ+

r (h) = Φ+r ∩ Φr(h). If X ∈ t + C(Φ+

r ) then X/2 ∈ t +C(Φ+

r (h)). Thus X → e<−ρ−µ,X/2>Ψ(hexp(X/2)) extends to a real analyticfunction on t + C(Φ+

r ). Therefore

(14.15) X →∑

h∈X(H,et)

Γ(h, t)e<−ρ−µ,X/2>Ψ(hexp(X/2))

extends to a real analytic function on t + C(Φ+r ).

By Lemma 14.5, for all X ∈ h such that texpX ∈ H ′,

(14.16) Ψ(texpX) = c−1∑

h∈X(H,et)

Γ(h, t)e<−ρ−µ,X/2>Ψ(hexp(X/2)).

By (14.15) there is an open neighborhood U of t in tT 0 such that Ψ extendsto be real analytic on U exp(C(Φ+

r )). �

We now turn to condition (C3). Fix α ∈ Φr and let J be the Cayleytransform of H as in Section 4. Choose c ∈ Aut(g(C)) satisfying (4.14)(a)and (b), and let β = c∗(α).

Lemma 14.17 Assume that Φ+ is a special set of positive roots, α ∈ Φ+,and that Φ+

J = c∗Φ+ is also special. Then α is a simple root for Φ+r .

Proof. A basic property of Cayley transforms is that for any root γ of H ,σ(c∗γ) = c∗(sασγ). Suppose γ ∈ Φ+

r , γ 6= α. Then

(14.18) σ(c∗γ) = c∗(sαγ).

If 〈γ, α∨〉 = 0 then sα(γ) = γ ∈ Φ+r . If 〈γ, α∨〉 6= 0 then c∗γ ∈ Φ+

J , and by(14.18) c∗γ is complex. Since Φ+

J is special σ(c∗γ) ∈ Φ+J , and so by (14.18)

again sαγ ∈ Φ+r . Therefore sαγ ∈ Φ+

r for all α 6= γ ∈ Φ+r , so α is simple for

Φ+r . �

Proposition 14.19 Suppose Φ+ is a special set of positive roots for H, suchthat Φ+

J is a special set of positive roots for J . Let g ∈ H(α). Then

[Dρ,+α∨ −Dρ,−

α∨ ]Ψ(H,Φ+, g) = 2DρJ

β∨Ψ(J ,Φ+J , g).

That is Ψ(H,Φ+) satisfies condition (C3).

70

Proof. Define Γ(H,Φ+) and Γ(J,Φ+J ) as in Definition 6.35.

By (14.7) we have

[Dρ,+α∨ −Dρ,−

α∨ ]Ψ(H,Φ+, g) =1

2c

∑

h∈X(H,eg)

Γ(H,Φ+, h, g)[Dρ,+α∨ −Dρ,−

α∨ ]Ψ(H,Φ+, h).

If h ∈ X(H, g) then for all γ ∈ Φ by (3.10) we have e2γ(h) = eγ(g). Thuseα(h) = ±1 and eγ(h) 6= ±1 for all γ 6= ±α. Thus X(H, g) = X(H, g, α)+ ∪X(H, g, α)− where

(14.20) X(H, g, α)± = {h ∈ X(H, g) : eα(h) = ±1}.

Suppose that h ∈ X(H, g, α)−. Then h is regular so that

(14.21) Dρ,+α∨ Ψ(H,Φ+, h) = Dρ,−

α∨ Ψ(H,Φ+, h) (h ∈ X(H, g, α)−).

On the other hand if h ∈ X(H, g, α)+, since Ψ(H,Φ+) satisfies (C3),

(14.22) [Dρ,+α∨ −Dρ,−

α∨ ]Ψ(H,Φ+, h) = 2DρJ

β∨Ψ(J,Φ+J , h)

where β = c∗α. Therefore(14.23)(a)

[Dρ,+α∨ −Dρ,−

α∨ ]Ψ(H,Φ+, g) = c−1∑

h∈X(H,eg,α)+

Γ(H,Φ+, h, g)DρJ

β∨Ψ(J,Φ+J , h).

On the other hand, by (14.7) applied to J

(14.23)(b) 2DρJ

β∨Ψ(J ,Φ+J , g) = c−1

∑

b∈X(J,eg)

Γ(J,Φ+J , b, g)D

ρJ

β∨Ψ(J,Φ+J , b).

Write X(J, g) = X(J, g, β)+ ∪X(J, g, β)− as above, so we have

(14.23)(c)

2DρJ

β∨Ψ(J ,Φ+J , g) = c−1

∑

b∈X(J,eg,β)+

Γ(J,Φ+J , b, g)D

ρJ

β∨Ψ(J,Φ+J , b)

+ c−1∑

b∈X(J,eg,β)−

Γ(J,Φ+J , b, g)D

ρJ

β∨Ψ(J,Φ+J , b)

Suppose h ∈ X(J, g, β)+. Then h ∈ H, so h ∈ X(H, g, α)+, andX(J, g, β)+ = X(H, g, α)+. Also by Lemma 6.37 Γ(H,Φ+, h, g) = Γ(J,Φ+

J , h, g).Therefore

71

(14.23)(d)

2DρJ

β∨Ψ(J ,Φ+J , g) = c−1

∑

b∈X(H,eg,α)+

Γ(H,Φ+, b, g)DρJ

β∨Ψ(J,Φ+J , b)

+ c−1∑

b∈X(J,eg,β)−

Γ(J,Φ+J , b, g)D

ρJ

β∨Ψ(J,Φ+J , b)

It is enough to show the second sum on the right hand side is 0, for then theresult follows from (14.23)(a) and (14.23)(d). We restate this as:

Lemma 14.24 Suppose α is an imaginary noncompact root and g ∈ H(α).Then

(14.25)∑

h∈X(H,eg,α)−

Γ(h, g)Dρα∨Ψ(H,Φ+, h) = 0.

Proof. Since H and Φ+ are fixed let Ψ(h) = Ψ(H,Φ+, h). Let w be the

reflection in α and suppose h ∈ H′. Since α is imaginary ǫR(Φ+, wh) =

ǫR(Φ+, h), and since Θ is stable Θ(wh) = Θ(h). Furthermore since ǫ(w) =−1,

∆0(Φ+, wh) = −eρ−wρ(h)∆0(Φ+, h).

ThusΨ(wh) = −eρ−wρ(h)Ψ(h).

Now assume h ∈ X(g, α)−, so eα(h) = −1. Then

(14.26) wh = hexp(πiα∨).

Therefore φ(wh) = p(g) and eα(wh) = −1, so wh ∈ X(g, α)−. Also, for allt ∈ R,

(14.27) e<ρ,itα∨>Ψ((wh)exp(itα∨)) = −eρ−wρ(h)e<ρ,−itα∨>Ψ(hexp(−itα∨)).

It follows from this and the definitions that

(14.28) Dρα∨Ψ(wh) = eρ−wρ(h)Dρ

α∨Ψ(h).

From (14.26) we have eρ−wρ(h) = (−1)〈ρ,α∨〉 and

(14.29) Γ(wh, g) =χ(g)

χ(wh)=

χ(g)

χ(h)χ(exp(πiα∨))= Γ(h, g)(−1)〈dχ,α

∨〉,

72

where (χ, χ) ∈ S(H,Φ+, χs, χs). Thus

(14.30) Γ(wh, g)Dρα∨Ψ(H,wh) = (−1)<dχ+ρ,α∨>Γ(h, g)Dρ

α∨Ψ(H, h).

By (6.2)

〈dχ+ ρ, α∨〉 = 〈2dχ+ 2ρ, α∨〉 ≡ 〈2dχ, α∨〉 (mod 2).

By [3, Section 7] 〈2dχ, α∨〉 = 1 (mod 2). Therefore

(14.31) Γ(wh, g)Dρα∨Ψ(H,wh) = −Γ(h, g)Dρ

α∨Ψ(H, h).

Therefore if wh 6= h then the two terms cancel, and if wh = h, Dρα∨Ψ(h) = 0.

�

This completes the proof of Proposition 14.19. �

Theorem 14.1 follows from Theorem 13.5, Propositions 14.4, 14.13 and14.19.

Remark 14.32 The only place that stability is used in the proof of Theorem14.1 is in the proof of Lemma 14.24.

It is easy to see that lifting preserves the property of being relativelytempered and supertempered, and (provided µ ∈ iz∗) tempered.

Proposition 14.33 Let Θ be a stable invariant eigendistribution on G.

1. If Θ is relatively tempered then LifteGGΘ is relatively tempered.

2. If Θ is relatively supertempered then LifteGGΘ is relatively supertempered.

3. Assume µ ∈ iz∗. If Θ is tempered then LifteGGΘ is tempered.

Proof. Since G and G have the same Lie algebra, we can use the sameEuclidean norm ‖ · ‖ on p to define both τH and τ eG.

Assume Θ is tempered and µ ∈ iz∗. Let h be a Cartan subalgebra of g.Fix g ∈ H ′ and write g = texpX where t ∈ T ′, X ∈ a. Then τ eG(g) = ‖X‖.Further, for any h ∈ H such that φ(h) = p(g), we have h = texpX/2 wheret ∈ T with φ(t) = p(t). Thus τH(h) = ‖X/2‖ = τ eG(g)/2.

73

Now, using (7.8)(d) and (13.7) applied to G, there are C, r ≥ 0 such that

|D(g)| 12 |Θ(g)| ≤ c−1∑

h∈X(Heg)

|D(g)| 12 |∆ eGG(h, g)||Θ(h)|

= c−1∑

h∈X(Heg)

|D(h)| 12 |Θ(h)| ≤ c−1[H1]C(1 + τ eG(g)/2)r,

where H1 = {h ∈ H : φ(h) = 1}. Thus there is a constant C ′ such that

|D(g)| 12 |Θ(g)| ≤ C ′(1 + τ eG(g))r

for all g ∈ H ′. Now Θ is tempered using (13.7) applied to G. This proves(3).

Part (1) is similar, using the fact that by (7.8)(c) |D(g)| 12 |∆ eGG(h, g)| =

|D(h)| 12 for any choice of (χs, χs) provided g ∈ 0G. Part (2) is also similar,using (13.8) in place of (13.7). �

15 Cuspidal Levi Subgroups

Let (G, G,G) be an admissible triple, and fix a cuspidal Levi subgroup M of

G. Let M = p−1(M), and M = p(M). These are cuspidal Levi subgroups of

G and G respectively.

Proposition 15.1 (M,M,M) is an admissible triple (Definition 3.14).

Proof. It is well known that M is the set of real points of the connectedreductive complex group M(C) and that Md(C) is acceptable, and it is clearthat every simple factor of M(C) is oddly laced. Let ΦM be the set of rootsof H(C) in M(C) where H = TA is a relatively compact Cartan subgroup

of M . Since G is an admissible cover of G, every noncompact root in ΦM ismetaplectic. Thus p : M → M is also an admissible cover. Further, M isthe set of real points of M(C) = M(C)/C, where C ⊂ Z0(G) ⊂ Z0(M) withc2 = 1 for all c ∈ C. This leaves only condition 3(c) of Definition 3.14.

Let Φ+ be a special set of positive roots forH in G and let Φ+M = Φ+∩ΦM .

Let ρ = ρ(Φ+) and ρM = ρ(Φ+M). Then ρ−ρM is zero on t. But C∩M0

d ⊂ T 0,so that eρ(c) = eρM (c) for all c ∈ C ∩M0

d . Thus if χ is a genuine character

of Z(M), then χ2(c)eρM (c) = χ2(c)eρ(c) = 1 for all c ∈ C ∩M0d . �

74

Suppose H is an arbitrary Cartan subgroup of M . Let Φ and ΦM be thesets of roots of H(C) in G(C) and M(C), respectively. Let Φ+ be a specialset of positive roots of Φ, and let Φ+

M = Φ+ ∩ ΦM . This is a special set ofpositive roots of ΦM . Suppose χ is a character of H . Define

(15.2) χM(h) = |eρM−ρ(h)|χ(h) (h ∈ H).

To be precise we are writing |eρM−ρ(h)| for |e2ρM−2ρ(h)| 12 . Note that Φi ⊂ ΦM ,so 2ρM −2ρ is a sum of real and complex roots; since Φ+ is special this takes

real values on h. Therefore ρM − ρ exponentiates to H0, and

(15.3) |eρM−ρ(h)| = eρM−ρ(h) (h ∈ H0).

In particular let Hs be a maximally split Cartan subgroup of M . Thisis also a maximally split Cartan subgroup of G. Let Φ+

s be a special set ofpositive roots for Hs(C) in G(C). Let ΦM,s be the roots of Hs(C) in M(C),and let Φ+

M,s = Φ+s ∩ ΦM,s.

Lemma 15.4 Fix (χs, χs) ∈ S(Hs,Φ+s ).

(1) Let χM,s

= (χs)M (cf. 15.2). Then (χs, χM,s) ∈ S(Hs,Φ

+M,s).

(2) Let H be any Cartan subgroup ofM and suppose (χ, χ) ∈ S(H,Φ+, χs, χs).Then (χ, χM) ∈ S(H,Φ+

M , χs, χM,s).

(3) Let µ = µ(χs, χs) and µM = µ(χs, χM,s) (defined with respect to M).Then µ = µM .

Proof. Fix h ∈ (Hs ∩Md)0. By (15.3) χM,s(h) = χs(h)e

ρM−ρ(h). On theother hand since (Hs ∩ Md)

0 ⊂ (Hs ∩ Gd)0, by (6.2)(a) we have χs(h) =

(χ2eρ)(h). Therefore

(15.5) χM,s(h) = χs(h)eρM−ρ(h) = (χ2eρ)(h)eρM−ρ(h) = (χ2

seρM )(h).

Therefore (6.2)(a) holds.Next let Γr(G,H) (resp. Γr(M,H)) be the group Γr(H) with respect

to the group G (resp. M) (cf. (2.6)). Then Γr(M,H) ⊂ Γr(G,H). Fixh ∈ Γr(M,H). Then |eρM−ρ(h)| = 1, and by Lemma 6.45

(15.6) χM,s(h) = χs(h) = ζcx(G,H)(h) = ζcx(M,H)(h).

This verifies condition (6.2)(b), and proves (1).

75

Now consider (2). Let χ0 be the character of H such that (χ, χ0) ∈S(H,Φ+

M , χs, χM,s) (cf. Proposition 6.31 and Definition 6.35). We need to

show χ0 = χM . We may assume A ⊂ As. Recall H = Γ(H)Z(G)H0

d. Firstassume h ∈ Γ(H)Z(G) ⊂ Γ(Hs)Z(G). Then |eρM−ρ(h)| = |eρM,s−ρs(h)| = 1,so χM(h) = χ(h) and χM,s(h) = χs(h). Therefore

(15.7)

χ0(h) = (χ/χs)(φ(h))χM,s(h) (by (6.34))

= (χ/χs)(φ(h))χs(h)

= χ(h) (by (6.34))

= χM(h).

Now suppose h = (H ∩Md)0 ⊂ H

0

d. Then χ(h) = (χ2eρ)(h), and

(15.8) χ0(h) = (χ2eρM )(h) = (χe−ρeρM )(h) = χM(h).

Next let X ∈ hd ∩ zM where zM is the center of m. Write h = expX.Then:

(15.9)

χ0(h) = (χ/χs)(φ(h))χM,s(h)

= (χ/χs)(φ(h))χs(h)eρM,s−ρs(h)

= (χ/χs)(φ(h))(χ2seρs)(h)eρM,s−ρs(h)

= (χ/χs)(exp2X)(χ2s)(expX)e<ρs,X>e<ρM,s−ρs,X>

But (χ/χs)(exp2X) = (χ/χs)2(expX). Also < ρM,s, X >=< ρM , X >= 0

since X ∈ zM . Therefore

(15.10)

χ0(h) = χ2(expX)

= χ(h)e−<ρ,X> (by (6.2)(a))

= χ(h)e<ρM−ρ,X> = χM (h).

The result follows from the fact that H0

d = exp(hd ∩ zM)(H ∩Md)0.

For the final assertion µ = dχs − 2dχs− ρ and µM = dχM,s− 2dχs − ρM ,and the result follows immediately from (15.2). �

Now fix lifting data (χs, χs) for G (Definition 7.1.) By Part (1) of the

Lemma (χs, χM,s) is lifting data for (M,M,M), and we use it to define

transfer factors ∆fMM

and LiftfMM . The remainder of this section is devoted to

proving that with these choices, lifting commutes with parabolic induction.

76

Suppose (h, g) ∈ X(H, H). We need to compare the transfer factorsdefined for G and for M . Recall

∆eGG(h, g) =

ǫr(h,Φ+)∆0(Φ+, h)χ(g)

ǫr(g,Φ+)∆0(Φ+, g)χ(h)(15.11)(a)

∆fMM

(h, g) =ǫr(h,Φ

+M )∆0(Φ+

M , h)χ(g)

ǫr(g,Φ+M)∆0(Φ+

M , g)χM(h).(15.11)(b)

Lemma 15.12 Suppose (h, g) ∈ X ′(H, H). Then

(15.13) ∆eGG(h, g)

|D eG(g)| 12|DG(h)| 12

= ∆fMM

(h, g)|DfM(g)| 12|DM(h)| 12

.

Proof. We first expand the left hand side, using (15.11)(a) and (5.1)(d) for

the |D| 12 terms. Let Φ+ be a special set of positive roots. The result is

(15.14)ǫr(h,Φ

+)∆0(Φ+, h)|∆0(Φ+, g)||eρ(g)|χ(g)

ǫr(g,Φ+)∆0(Φ+, g)|∆0(Φ+, h)||eρ(h)|χ(h).

Write Φ+r ,Φ

+i and Φ+

cx for the real, complex and imaginary roots in Φ+,respectively. Then with the obvious notation

(15.15) ∆0(h,Φ+) = ∆0(h,Φ+r )∆0(h,Φ+

cx)∆0(h,Φ+

i ).

Then

(15.16)ǫr(h,Φ

+)∆0(Φ+r , h) = |∆0(Φ+

r , h)|∆0(Φ+

cx, h) = |∆0(Φ+cx, h)|

Therefore

(15.17)ǫr(h,Φ

+)∆0(Φ+, h)|∆0(Φ+, g)|ǫr(g,Φ+)∆0(Φ+, g)|∆0(Φ+, h)| =

∆0(Φ+i , h)|∆0(Φ+

i , g)|∆0(Φ+

i , g)|∆0(h,Φ+i )|

.

A similar argument applies to the right hand side of (15.13). Since Φi = ΦM,i

and the right hand side of (15.17) only depends on the imaginary roots, weconclude

(15.18)ǫr(h,Φ

+)∆0(Φ+, h)|∆0(Φ+, g)|ǫr(g,Φ+)∆0(Φ+, g)|∆0(Φ+, h)| =

ǫr(h,Φ+M)∆0(Φ+

M , h)|∆0(Φ+M , g)|

ǫr(g,Φ+M)∆0(Φ+

M , g)|∆0(Φ+M , h)|

77

It remains to show

(15.19)|eρ(g)|χ(g)

|eρ(h)|χ(h)=

|eρM (g)|χ(g)

|eρM (h)|χM(h).

Recall χM(h) = χ(h)|eρM−ρ(h)|. Inserting this it suffices to show

(15.20) |eρ−ρM (g)| = |eρ−ρM (h)|2.

which follows from the fact that |e2ρ−2ρM (h)| = |eρ−ρM (φ(h))| = |eρ−ρM (g)|.�

We can rewrite the Lemma in terms of orbits.

Lemma 15.21 Suppose OfM ∈ Orb(M) and Ost

M∈ Orbst(M) are strongly

regular and semisimple, and satisfy φ(Ost

M) = p(OfM)st.

Let O eG be the unique G-orbit containing OfM and let Ost

Gbe the unique

stable G-orbit containing Ost

M. Then φ(Ost

G) = p(O eG)st, and

∆eGG(Ost

G, O eG)

|D eG(O eG)| 12|DG(Ost

G)| 12

= ∆fMM

(Ost

M,OfM)

|DfM(OfM )| 12|DM(Ost

M)| 12

.

Lemma 15.22 Suppose O eG ∈ Orb(G) is strongly regular and Ost

M∈ Orbst(M).

Let Ost

Gbe the unique stable orbit of G containing Ost

M. Then

(15.23) φ(Ost

G) = p(O eG)st ⇔ φ(Ost

M) = p(OfM)st

for some M -orbit OfM ⊂ O eG. Furthermore, if O eG is relevant, then OfM isunique.

O eG

��

OfM?_oo_ _ _ _ _

��p(O eG)st p(OfM )st?

_oo_ _ _

OstG

φ

OO

OstM

?_oo

φ

OO��

�

78

Proof. Choose h ∈ M , g ∈ G so that OstM

= Ost(M,h) and O eG = O(G, g).

Then OstG

= Ost(G, h). Let h = φ(h) and g = p(g). Note that p(O eG) =O(G, g) and φ(Ost

M) = Ost(M,h).

Except for the final statement the result does not involve G or M , andsays

(15.24) Ost(G, h) = Ost(G, g) ⇔ Ost(M,h) = Ost(M, y)

for some y ∈ O(G, g) ∩M . The implication ⇐ is obvious: y ∈ O(G, g) andy ∈ Ost(M,h) implies Ost(G, g) = Ost(G, y) = Ost(G, h).

Suppose Ost(G, h) = Ost(G, g). By [24, Section 2]

(15.25) Ost(G, h)/G ≃ Ost(M,h)/M.

More precisely if Ost(M,h) = ∪iO(M,hi) with hi ∈ M , then Ost(G, h) =∪iO(G, hi). Therefore O(G, g) = O(G, hi) for some hi ∈ M . Take y = hi.This proves the implication ⇒.

Assume that g is relevant. By the preceding paragaraph clearly Ost(M,h)contains a unique M-orbit of the form O(M, y) with y G-conjugate to G.

Choose y ∈ p−1(y) so that y is G-conjugate to g, i.e. O(M, y) ⊂ O eG. It is

clear that φ(OstM

) = p(O(M, y))st, and the only other choice is O(M,−y). By

Lemma 9.4(3), since y is relevant, y is not G-conjugate to −y, so O(M,−y)is not contained in O eG. �

We continue to work with our given Levi subgroups M and M . Let MNand MN be parabolic subgroups of G and G with Levi components M andM respectively. Let ΘM be a stable character of M and let ΘfM be a character

of M . Then the induced characters are independent of the choices of N andN so by abuse of notation we write

(15.26) IndGM

(ΘM) = IndGMN

(ΘM ⊗ 1), IndeGfM

(ΘfM) = IndeGfM eN

(ΘfM ⊗ 1).

Theorem 15.27 Let ΘM be a stable character of M . Then

(15.28) IndeGfM

(LiftfMM

(ΘM)) = LifteGG(IndG

M(ΘM)).

Proof. Suppose O eG ∈ Orb(G) is strongly regular and semisimple. Let

ReseGfMO eG = {OfM | OfM ⊂ O eG}. Then for any character ΘfM of M , the induced

79

character is given by (for example see [13])

(15.29) IndeGfM

(ΘfM)(O eG) = |D eG(O eG)|− 12

∑

OfM∈Res

eGfM

(O eG)

|DfM(OfM)| 12 ΘfM(OfM).

Therefore by (9.11)

(15.30)

IndeGfM

(LiftfMM

(ΘM))(O eG) =∑

OfM

|DfM(OfM)| 12|D eG(O eG)| 12

LiftfMM

(ΘM)(OfM)

= c−1M

∑

OfM

∑

Ost

M

|DfM(OfM)| 12|D eG(O eG)| 12

∆fMM

(OstM,OfM)ΘM(Ost

M).

The double sum is over

(15.31) {(OfM ,OstM

) | OfM ∈ ReseGfM

(O eG), φ(OstM

) = p(OfM)st}

Similarly, suppose OstG∈ Orbst(G) is a semisimple orbit and let ResG

M(Ost

G) =

{OstM| Ost

M⊂ Ost

G}. It is well known that IndGM(ΘM) is stable, and its has

character formula

(15.32) IndGM(ΘM)(OstG) = |DG(Ost

G)|− 1

2

∑

Ost

M∈ResG

M(Ost

G)

|DM(OstM

)| 12 ΘM(OstM

).

This follows from the previous induced character formula and (15.25). There-fore

(15.33)

LifteGG(IndGM(ΘM)(O eG) = c−1

G

∑

Ost

G

∆eGG(Ost

G,O eG)IndGM(ΘM))(Ost

G)

= c−1G

∑

Ost

G

∑

Ost

M

|DM(OstM

)| 12|DG(Ost

G)| 12

∆eGG(Ost

G,O eG)ΘM(Ost

M).

In this case the double sum is over

(15.34) {(OstG,Ost

M) |φ(Ost

G) = p(O eG)st,Ost

M∈ ResG

M(Ost

G)}

Suppose that O eG is not relevant. Then (15.33) is zero since there are no

orbits OstG

satisfying φ(OstG) = p(O eG)st. Then for all OfM ∈ Res

eGfM

(O eG), OfM

is not relevant, so (15.30) is also zero.

80

Assume O eG is relevant. Since a maximally split Cartan subgroup Hs ofM is also one for G, cM = cG. By Lemma 15.22 both sums are over the sameset of orbits Ost

M. For each such orbit, the equality of the corresponding terms

is then given by Lemma 15.21. �

By Theorem 15.27, we see that to understand lifting of standard represen-tations of G, it suffices to understand lifting of discrete series representationsof its cuspidal Levi subgroups.

16 Modified Character Data

We describe data which parametrizes L-packets for G, and also data forirreducible representations of G. These are provided by character data ofVogan. For G see [26, Definition 6.6.1], and for G see [28, Section 2]. Forour purposes it is convenient to use a modified version of this data.

First let G be the real points of a connected, complex reductive group.Following [26, Definition 6.6.1] we define a regular character ofG to be a tripleγ = (H,Γ, λ) consisting of a θ-stable Cartan subgroup H of G, a characterΓ of H , and an element λ ∈ h(C)∗. We assume 〈λ, α∨〉 ∈ R× for all α ∈ Φi.Define Φ+

i = {α ∈ Φi | 〈λ, α∨〉 > 0} and let Φ+i,c = Φi ∩ Φi,c (cf. Section 2).

Define ρi = 12

∑Φ+

iα and ρi,c = 1

2

∑Φ+

i,cα. We assume

(16.1) dΓ = λ+ ρi − 2ρi,c.

Write H = TA and let M = CentG(A) as usual. Let P be any parabolicsubgroup containing M . Associated to γ = (H,Γ, λ) is a relative discreteseries representation πM = πM(γ) of M , and a standard module πG(γ) =π(γ) = IndGP (πM).

The central character of πM (γ) (respectively π(γ)) is Γ|Z(M) (resp. Γ|Z(G)).The Harish-Chandra parameter of πM(γ) is λ, and the infinitesimal characterof πM(γ) and π(γ) is λ.

The definition of character data is designed to make the lowest K-typesof π(γ) evident. For example if H is a compact Cartan subgroup then Γ isthe highest weight of a lowest K-type of π(γ). We are interested in char-acter formulas, so our needs are somewhat different. We modify this dataappropriately.

81

Definition 16.2 A modified regular character for G is a triple γ = (H,Γ, λ)where H is a θ-stable Cartan subgroup of G, Γ is a character of H, andλ ∈ h(C)∗. We assume

(16.4) 〈λ, α∨〉 ∈ R× for all α ∈ Φi.

Let Φ+i (λ) = {α ∈ Φi | 〈λ, α∨〉 > 0} and ρi(λ) = 1

2

∑α∈Φ+

i (λ) α. We assume

(16.5) dΓ = λ− ρi(λ).

Let CD(G) be the set of modified regular characters. If H is fixed letCD(G,H) be the set of modified regular characters (H,Γ, λ). If H is givenwe write (Γ, λ) = (H,Γ, λ).

Write H = TA and let M = CentG(A). Note that Φi is the set of roots ofh(C) in m(C). Since the definition of modified regular characters only refersto the imaginary roots, CD(G,H) = CD(M,H).

Associated to γ is a relative discrete series representation πM (γ), withHarish-Chandra parameter λ and central character Γ|Z(M). Let ΘM(γ) bethe character of πM (γ). Then for h ∈ H ′,(16.6)

ΘM(γ)(h) = (−1)qM ∆0(Φ+i (λ), h)−1

∑

w∈W (M,H)

ǫ(w)ewρi(λ)−ρi(λ)(h)Γ(w−1h)

where qM = 12dim(Md/K ∩Md) and ∆0 is given by (5.1).

Let P = MN be any parabolic subgroup containingM , and define π(γ) =πG(γ) = IndGP (πM(γ)). Since we are interested only in characters the choiceof P is not important. Let ΘG(γ) be the character of πG(γ):

(16.7) ΘG(γ) = IndGP (ΘM(γ)).

Suppose γ = (H,Γ, λ) is a modified character. Note that (with the ob-vious notation) 2ρi(λ) − 2ρi,c(λ) is a sum of roots, so e2ρi(λ)−2ρi,c(λ) is a welldefined character of H(C), and by restriction of H . This character is trivialon Z(M). It follows easily that (H,Γe2ρi(λ)−2ρi,c(λ), λ) is a regular characterin the sense of Vogan, and defines the same relative discrete series represen-tation of M and standard representation of G. This construction is clearlya bijection between modified character data and character data in the senseof Vogan.

There is a natural action of G on CD(G) by conjugation. The precedingbijection is G-equivariant. By [26] or [28, Theorem 2.9] we conclude:

82

Lemma 16.8 Suppose γ, γ′ ∈ CD(G). Then ΘG(γ) = ΘG(γ′) if and only ifγ = gγ′ for some g ∈ G.

We don’t obtain every irreducible representation of G this way, sincewe don’t allow “limit” characters, but this won’t matter since we are onlyinterested in stable virtual characters. Note that Wi = W (Φi) acts on Hand this induces an action on CD(G,H): w(H,Γ, λ) = (H,wΓ, wλ) wherewΓ(h) = Γ(w−1h). Note that πG(wγ) and πG(γ) have the same infinitesimaland central character.

Definition 16.9 Suppose γ = (H,Γ, λ) is a modified regular character. Let

(16.10) πst

M(γ) =∑

w∈W (M,H)\Wi

πM(wγ).

and

(16.11) πst

G(γ) =∑

w∈W (M,H)\Wi

πG(wγ).

Let ΘstM(γ) and Θst

G(γ) be the characters of πstM(γ) and πst

G(γ), respectively.

The character formula for ΘstM(γ) onH is the same as (16.6) withW (M,H)

replaced by Wi.

Lemma 16.12 ([24], Lemma 5.2) Fix γ ∈ CD(G,H). Then ΘstM(γ) and

ΘstG(γ) are stable characters.

Suppose Θ is a stable virtual character of G. Then there exist γ1, . . . , γn ∈CD(G) and integers a1, . . . , an so that Θ =

∑aiΘ

G,st(γi).

Stable characters were defined before Definition 9.10. While the secondpart of the lemma is not stated in [24] it follows easily from the proof. SeeDefinition 18.9 and Lemmas 18.10 and 18.11 of [7].

Now let G be an admissible cover of G. The definition of regular characterextends naturally to G. See [11, Section 27] and [28, Section 2]. In this setting

Γ is an irreducible representation of H, or (by Lemma 10.8) a character of

Z(H).

Definition 16.13 A genuine modified regular character of G is a triple(H, Γ, λ). Here H is a Cartan subgroup of G, Γ is a genuine one-dimensional

83

representation of Z(H), and λ ∈ h(C)∗. As in Definition 16.2. we require

〈λ, α∨〉 ∈ R× for all α ∈ Φi, and dΓ = λ − ρi(λ). Let CDg(G) be the set

of genuine modified regular characters of G, and CDg(G, H) the subset with

given Cartan subgroup H.

Let H = TA be the image of H in G, M = CentG(A), and M = p−1(M).

Associated to γ = (H, Γ, λ) is a relative discrete series representation πfM (γ)

of M . The formula for the character ΘfM(γ) is (cf. 16.6)(16.14)

ΘfM(γ)(h) = (−1)qM ∆0(Φ+i (λ), h)−1

∑

w∈W (fM, eH)

ǫ(w)ewρi(λ)−ρi(λ)(h)Tr(τ (Γ)(w−1h))

for h ∈ H ′. Here τ(Γ) is the irreducible representation of H associated to Γby Lemma 10.8.

As before we also associate to γ the standard module π eG(γ) for G induced

from πfM(γ). The central character of πfM(γ) is the restriction of Γ to the

center of M , and similarly for π eG(γ).

Lemma 16.15 Suppose γ, γ′ ∈ CDg(G). Then π eG(γ) ≃ π eG(γ′) if and only

if γ′ = gγ for some g ∈ G. Every irreducible genuine representation of G isisomorphic to π eG(γ) for some γ ∈ CDg(G).

Proof.The first statement is a special case of [28, Theorem 2.9].

The second statement amounts to the fact that M has no genuine limitsof (relative) discrete series. This is because every irreducible genuine rep-

resentation of G is of the form π eG(γ) where γ is final limit data as in [28,Definition 2.4]. For final limit data we allow 〈λ, α∨〉 = 0 if α is a noncompact

imaginary root. But if Γ is genuine, then by [3, Lemma 6.11] 〈dΓ, α∨〉 ∈ Z+ 12

for all noncompact imaginary roots. This also holds for λ, so 〈λ, α∨〉 6= 0 forall noncompact imaginary roots. Therefore every genuine final limit data forG is in fact regular. �

There is no natural definition of stable distribution for G, and we do notdefine analogues of πst

G(γ) and πstG(γ) for G.

We will make frequent use of formal sums of modified regular characters(for example in Lemma 16.12). So if γi ∈ CD(G) and ai ∈ Z (i ≤ n) we

define ΘG(∑

i aiγi) =∑

i aiΘG(γi), and similar notation applies to G.

84

17 Formal Lifting of Modified Character Data

Fix an admissible triple (G, G,G). Choose a maximally split Cartan sub-group Hs of G, and (χs, χs) ∈ S(Hs). This data determines transfer factors

for G and G (Section 7) and LifteGG

is defined (Section 9). Let µ = µ(χs, χs) ∈z∗ as in (6.36)(c).

Suppose H is a θ-stable Cartan subgroup of G and γ = (H,Γ, λ) ∈CD(G,H) (Definition 16.2). We define the lift of γ to CDg(G, H). We first

choose transfer factors for lifting from H to H. These differ from the obviouschoice by a ρ-shift.

Write H = TA and let M = CentG(A) as usual. Choose an arbitrary

genuine character χ of Z(H) and a special set Φ+ of positive roots of H inG. We assume

(17.1) Φ+ ∩ Φi = {α ∈ Φi | 〈λ, α∨〉 > 0}.

Let χ0 be the character of H so that (χ, χ0) ∈ S(H,Φ+, χs, χs) (Definition6.35). Define

(17.2) χ = |eρi−ρ|χ0.

Let φH be φ restricted to H , and use it to define

(17.3) LifteHH

(Γ) = LifteHH

(χ, χ,Γ)

(cf. 10.15).

Lemma 17.4 χ is independent of the choice of special positive roots Φ+

satisfying (17.1), and LifteHH is independent of the choice of χ.

Proof. This follows from Lemmas 6.15 and 6.47. �

Definition 17.5 If Γ|Ker(φH) 6= χ|Ker(φH) then LifteHH

(Γ) = 0, and we define

LifteGG(γ) = ∅ and Θ eG(Lift

eGG(γ)) = 0.

Otherwise write

(17.6) LifteHH

(Γ) =n∑

i=1

τ (Γi)

85

where τ(Γi) is given by Lemma 10.8, n = |p(Z(H))/φ(H)|, and each Γiis a genuine one-dimensional representation of Z(H). For 1 ≤ i ≤ n let

γi = (H, Γi,12(λ− µ)); this is a genuine regular character of G. Define

(17.7) LifteGG(γ) = {γ1, . . . , γn}.

and

(17.8) Θ eG(LifteGG(γ)) =

n∑

i=1

Θ eG(γi).

It is important to keep in mind that LifteHH

depends on χ, and thereforeon λ and γ (cf. Lemma 17.4).

The next Lemma follows from the definitions.

Lemma 17.9 Let g ∈ G and define g = p(p(g)) ∈ G. Then LifteGG(gγ) =

gLifteGG(γ).

The following Lemma is an easy consequence of Lemma 17.4 and Corol-lary 10.16.

Lemma 17.10 Fix γ = (H, Γ, λ) ∈ CDg(G). Then there is a unique γ ∈CD(G) such that γ occurs in Lift

eGG(γ). It is given by γ = (H,Γ, 2λ+µ) where

Γ is the unique character of H such that τ(Γ) occurs in LifteHH(Γ) (cf. Corol-

lary 10.16).

The reason for the shift (17.2) in the definition of LifteHH is that it makes

the next Lemma hold. Fix a cuspidal Levi subgroup M of G. Recall (Section

15) (M,M,M) is an admissible triple, where M = p−1(M) and M = p(M),

and LiftfMM was defined in Section 15 (following Lemma 15.4).

Lemma 17.11

(17.12) IndeGfM

(ΘfM(LiftfMM(γ))) = Θ eG(Lift

eGG(γ)).

Proof. Fix a genuine character χ of Z(H).Let χ0 be the unique character of H so that (χ, χ0) ∈ S(H,Φ+, χs, χs)

and set χ1 = χ0|eρi−ρ|. Let µ = µ(χs, χs) = dχ0 − 2dχ − ρ(Φ+). Then

86

the right hand side is the sum of terms IndeGfM

(ΘfM(H, Γ, 12(λ − µ)) where

Γ ∈ LifteHH(χ, χ1,Γ).

On the other hand let χ2 be the unique character of H so that (χ, χ2) ∈S(H,Φ+

M , χs, χM,s). Let µM = µ(χs, χM,s) = dχ2 − 2dχ − ρ(Φ+i ). Then (cf.

Lemma 15.4) the left hand side is the sum of terms IndeGfM

(ΘfM(H, Γ, 12(λ −

µM))) where Γ ∈ LifteHH(χ, χ2,Γ).

It is enough to show χ1 = χ2, i.e. (χ, χ0) ∈ S(H,Φ+, χs, χs) implies(χ, χ0|eρi−ρ|) ∈ S(H,Φ+

M , χs, χM,s). This is Lemma 15.4(2). �

Corollary 17.13 Suppose γ = (H, Γ, λ) ∈ CDg(G, H). Then there is a

character Γ of H such that γ = (H,Γ, 2λ + µ) ∈ CD(G,H), and γ is a

summand of LifteGG(γ).

Proof. This follows easily from Corollary 10.16: take Γ(h) = χ(h)(Γ/χ−1)(φ(h)).We leave the details to the reader. �

Given γ = (H,Γ, λ) ∈ CD(G,H) it is helpful to know when LifteGG(γ) is

non-zero for some choice of lifting data.

Lemma 17.14 We can choose lifting data so that LifteGG(γ) 6= 0 if and only

if

(17.15) Γ(h) =

{(χ2eρ)(h) h ∈ Ker(φH) ∩H0

d,

ζcx(h) h ∈ Γr(H).

Here, as usual, χ is any genuine character of H, and ρ is defined with respectto a special set of positive roots (Definition 2.1).

Proof. Fix lifting data (χs, χs), and let χ0 be the character of H such that(χ, χ0) ∈ S(H,Φ+, χs, χs). Then the lift is non-zero if Γ(h) = |eρi−ρ(h)|χ0(h)for all h ∈ Ker(φH). It is easy to see |eρi−ρ(h)| = 1 for all h ∈ Ker(φH), sothe condition is

(17.16)(a) Γ(h) = χ0(h) (h ∈ Ker(φH)).

Using (6.2) the condition in (17.15) is equivalent to

(17.16)(b) Γ(h) = χ0(h) (h ∈ Gd ∩ Ker(φH)).

87

This condition is obviously necessary; we need to show it is sufficient.Let λ be a character of H satisfying

λ(h) = 1 (h ∈ H ∩Gd)(17.16)(c)

λ(h) = Γ(h)χ−10 (h) (h ∈ Ker(φH)).(17.16)(d)

This is possible since H∩Gd∩Ker(φH) = Gd∩Ker(φH), and Γχ−10 = 1 on this

group by (b). Then λ extends uniquely to a one dimensional representationofHGd, and then (possibly not uniquely) to a one dimensional representationψ of G.

The result follows from Lemma 7.15: the lift is non-zero if we replace∆

eGG(χs, χs) with ψ∆

eGG(χs, χs). �

18 Lifting Stable Discrete Series

Assume that (G, G,G) is an admissible triple. Fix lifting data (χs, χs) (Def-

inition 7.1). This data determines choices of transfer factors for G and G

(Section 6) and LifteGG

is defined (Section 7). We assume for this section onlythat G has a relatively compact Cartan subgroup H , i.e. such that H ∩Gd iscompact, and hence G, G, and G have relative discrete series representations.We consider lifting of the relative discrete series from G to G. See Section12 for the case when G = G is connected and semisimple.

Fix a relatively compact Cartan subgroup H of G with roots Φ. Since His relatively compact Φ = Φi. Fix γ = (H,Γ, λ) ∈ CD(G,H), and let Θst

G(γ)

be the associated stable discrete series character as in Section 16. We wantto compute Lift

eGG(Θst

G(γ)).

Define LifteHH

(Γ) and LifteGG(γ) = (H,Lift

eHH

(Γ), 12(λ − µ)) as in Definition

17.5. In this case by (17.3)

(18.1) LifteHH

(Γ) = LifteHH

(χ, χ,Γ)

where (χ, χ) ∈ S(H,Φ+(λ), χs, χs) (since Φ = Φi, Φ+ is uniquely determinedand χ = χ0). Fix (χ, χ) ∈ S(H,Φ+(λ), χs, χs).

Let W = W (Φ) = Wi. For w ∈ W let wγ = (H,wΓ, wλ). Then

LifteHH(wγ) = (H,Lift

eHH(wΓ), 1

2w(λ − µ)) is defined. It is important to keep

88

in mind that the lifting used in defining LifteHH(wΓ) depends on w: by the

preceding discussion

(18.2) LifteHH

(wΓ) = LifteHH

(χw, χw, wΓ)

where (χw, χw) ∈ S(H,Φ+(wλ), χs, χs). One possibility is to take

(18.3) (χw, χw) = (χ, χewρ−ρ).

To see this, write h = γh0 where γ ∈ Γ(H) and h0 ∈ H0. By Lemma 6.47 we

can take χw = χ and χw(h) = χ(h)ewρ−ρ(h0). Since H is relatively compactΓ(H) ⊂ Z(G), so ewρ−ρ(h0) = ewρ−ρ(h).

We will also use the fact that if x ∈W (G, H), y ∈W then

(18.4) LifteHH

(xyΓ)(h) = LifteHH

(yΓ)(x−1h),

or more explicitly

(18.5) LifteHH

(χxy, χxy, xyΓ)(h) = LifteHH

(χy, χy, yΓ)(x−1h).

This follows from the fact we can take (χy, χy) as in (18.3) and (χxy, χxy) =(xχy, xχy). Then

(18.6) LifteHH(χxy, χxy, xyΓ)(h) = Lift

eHH(xχy, xχy, xyΓ)(h)

and it follows immediately from the definitions that this equals

(18.7) LifteHH(χy, χy, yΓ)(x−1h).

Proposition 18.8 Let γ ∈ CD(G,H). Then

(18.9) LifteGG(Θst

G(γ)) = CG(H)

∑

w∈W (G,H)\W

Θ eG(LifteGG(wγ)).

Proof. Let Θ = ΘstG(γ). By (16.6)

(18.10) Θ(h) = (−1)q∆0(Φ+, h)−1∑

w∈W

ǫ(w)ewρ−ρ(h)Γ(w−1h) (h ∈ H ′)

89

where q = qG. Fix a strongly regular element h ∈ H . Since H has no realroots we have

∆eGG(h, h) =

∆0(Φ+, h)χ(h)

∆0(Φ+, h)χ(h)(h ∈ X(H, h)).

By definition of lifting we have LifteGG(Θ)(h) =

(18.11) c−1(−1)q∆0(Φ+, h)−1∑

w∈W

ǫ(w)χ(h)∑

h∈X(H,eh)

χ(h)−1ewρ−ρ(h)wΓ(h).

If h ∈ X(H, h) then ewρ−ρ(h) = e2wρ−2ρ(h), so ewρ−ρ(h) = ewρ−ρ(h)eρ−wρ(h).Therefore

(18.12)

LifteGG(Θ)(h) = c−1(−1)q∆0(Φ+, h)−1

∑

w∈W

ǫ(w)ewρ−ρ(h)×

χ(h)∑

h∈X(H,eh)

χ(h)−1eρ−wρ(h)wΓ(h).

By 10.2

(18.13) χ(h)∑

h∈X(H,eh)

χ(h)−1eρ−wρ(h)wΓ(h) = c(H)LifteHH(χ, χewρ−ρ, wΓ)(h).

By (18.3) LifteHH(χ, χewρ−ρ, wΓ) = Lift

eHH(wΓ). Recalling CG(H) = C(H) =

c(H)/c this gives(18.14)

LifteGG(Θ)(h) = C(H)(−1)q∆0(h,Φ+)−1

∑

w∈W

ǫ(w)ewρ−ρ(h)LifteHH(wΓ)(h)

Write the sum as

(18.15)∑

y∈W ( eG, eH)\W

ǫ(y)eyρ−ρ(h)∑

x∈W ( eG, eH)

ǫ(x)exyρ−yρ(h)LifteHH(xyΓ)(h).

By (18.5)

(18.16) LifteHH(xyΓ)(h) = Lift

eHH(yΓ)(x−1h).

90

Also, for all y ∈W we have

(18.17) ∆0(h,Φ+)−1ǫ(y)eyρ−ρ = ∆0(h, yΦ+)−1.

This gives

(18.18)

LifteGG(Θ)(h) = C(H)(−1)q

∑

y∈W ( eG, eH)\W

∆0(yΦ+, h)−1×

∑

x∈W ( eG, eH)

ǫ(x)exyρ−yρ(h)LifteHH(yΓ)(x−1h).

By Proposition 10.11 LifteHH(yΓ) = 0 unless y ∈W (γ) where

(18.19) W (γ) = {w ∈W |χ(h) = eρ−wρ(h)wΓ(h) for all h ∈ Ker(φH)}.

Fix y ∈W (γ) and define

(18.20) Γy(h) = χ(h)χ−1(h)eρ−yρ(h)yΓ(h) (h ∈ X(H, h)).

Then by Proposition 10.11

(18.21) LifteHH(yΓ)(h) =

∑

eΓ∈Xg( eH,eΓy)

Tr(τ(Γ)(h)) (h ∈ H).

Thus

(18.22)

∑

x∈W ( eG, eH)

ǫ(x)exyρ−yρ(h)LifteHH(yΓ)(x−1h)

=∑

eΓ∈Xg( eH,eΓy)

∑

x∈W ( eG, eH)

ǫ(x)exyρ−yρ(h)Tr(τ(Γ)(x−1h)).

But for each Γ ∈ Xg(H, Γy) by (16.14) we have(18.23)

(−1)q∆0(yΦ+, h)−1∑

x∈W ( eG, eH)

ǫ(x)exyρ−yρ(h)Tr(τ(Γ)(x−1h)) = Θ eG(γ)(h)

where γ = (H, Γ, (yλ−µ)/2) is the corresponding element of LifteGG(yγ). Thus

(18.24) LifteGG(Θ)(h) = C(H)

∑

y∈W ( eG, eH)\W (γ)

∑

eγ∈LifteG

G(yγ)

Θ eG(γ)(h).

91

In other words for every strongly regular element h ∈ H we have

(18.25) LifteGG(Θ)(h) = C(H)

∑

w∈W (G,H)\W

Θ eG(LifteGG(wγ))(h).

Since Θ is a stable discrete series character of G with Harish-Chandraparameter λ, it is a stable invariant eigendistribution with infinitesimal char-acter λ and is relatively supertempered. Thus by Theorems 10.32 and 10.35

LifteGG(Θ) is an invariant eigendistribution on G with infinitesimal character

(λ−µ)/2 and is relatively supertempered. Now each Θ eG(LifteGG(wγ)), w ∈W ,

is either zero or is a sum of discrete series characters which all have infinitesi-mal character (λ−µ)/2, and hence is also a relatively supertempered invariant

eigendistribution on G with infinitesimal character (λ− µ)/2. Thus (18.25)

holds for all g ∈ G′ by Harish-Chandra’s Theorem 13.9. �

Not all of the terms in (18.9) are non-zero. For a precise description ofwhich ones are non-zero see Proposition 19.15.

Example 18.26 Let G = SL(2,R). A discrete series representation of Gor G is determined by its Harish-Chandra parameter λ ∈ t∗(C) where T isa compact Cartan subgroup of G. Write πst

G(λ) for the corresponding stablesum of discrete series of G, and π eG(λ) for a discrete series representation of

G. Let P = Z〈ρ〉 be the weight lattice. We compute C(T ) = 2, and then

(18.27) LifteGG(πst

G(λ)) =

{2π eG(λ/2) + 2π eG(−λ/2) λ ∈ 2P + ρ

0 otherwise

In the usual coordinates the lift is non-zero if and only if λ is an odd integer,in which case ±λ/2 ∈ Z + 1

2.

If G = PSL(2,R) ≃ SO(2, 1) the result is the similar, except that C(T ) =1.

19 Lifting Standard Representations

Fix an admissible triple (G, G,G) and lifting data (χs, χs) for G.Suppose Θst

Gis a stable virtual character of G. Recall (Lemma 16.12) that

ΘstG

is a sum of standard characters. Therefore, assuming we can write ΘstG

as

92

a sum of standard modules, it is enough to compute the lifting of an arbitrarystandard character. This is the main result of this section (Theorem 19.1).

We recall some constructions from Section 16. Suppose γ is modifiedcharacter data for G (Definition 16.2), with associated stable standard char-acter Θst

G(γ). Write γ = (H,Γ, λ) and let M be the corresponding Levi factor

of G. Recall γ is also character data for M , so the stable relative discrete

series character ΘstM

(γ) is defined, and ΘstG(γ) = IndGM(Θst

M(γ)). (As in the

previous section we write IndGM

instead of IndGMN

).

Recall (Section 16) LifteGG(γ) is a set of modified character data for G.

Recall Wi = W (Φi) acts on CD(G,H).

Theorem 19.1

(19.2) LifteGG(Θst

G(γ)) = CG(H)

∑

w∈W (M,H)\Wi

Θ eG(LifteGG(wγ)).

Explicitly for w ∈Wi define LifteGG(wγ) by Definition 17.5, and define γ(w, i) ∈

CDg(G) by:

(19.3) LifteGG(wγ) =

nw∑

i=1

γ(w, i)

(if the lift is empty take nw = 0). Then

(19.4) LifteGG(Θst

G(γ)) = CG(H)

∑

w∈W (M,H)\Wi

nw∑

i=1

Θ eG(γ(w, i)).

Proof. This is merely a question of assembling the pieces. Write∑

w forthe sum over W (M,H)\Wi. Then(19.5)

LifteGG(Θst

G(γ)) = Lift

eGG(IndGM(Θst

M(γ))) ((16.7) and Definition (16.9)

= IndeGfM

(LiftfMM(Θst

M(γ))) (Theorem 15.27)

= CG(H)IndeGfM

(∑

w

ΘfM(LiftfMM(wγ)) (Proposition 18.8)

= CG(H)∑

w

Θ eG(LifteGG(wγ)) (Lemma 17.11).

93

�

It is important to know that the terms in the sum (19.4) are distinct.

Proposition 19.6 In the sum (19.4), π eG(γ(w, i)) ≃ π eG(γ(w′, i′)) if and onlyif W (M,H)w = W (M,H)w′ and i = i′.

Proof. Suppose for i = 1, 2, wi ∈ Wi, and γi occurs in LifteGG(wiγ). We

may as well assume w1 = 1, and write γ1 = (H, Γ1,12(λ − µ)) ∈ Lift

eGG(γ),

γ2 = (H, Γ2,12(wλ− µ)) ∈ Lift

eGG(wγ) for some w ∈Wi. We want to show

(19.7) (H, Γ2,1

2(wλ− µ) = g(H, Γ1,

1

2(λ− µ))

for some g ∈ G implies w = 1 and Γ1 = Γ2. Since g normalizes H let u bethe image of p(p(g)) in W (G,H) and set v = w−1u. Both w and u are inW θ, the elements of W fixed by θ, so v ∈W θ. Also uλ = wλ, so vλ = λ andtherefore vΦ+

i (λ) = Φ+i (λ).

We now apply [27, Propositions 3.12 and 4.16]. By Proposition 3.12(c)v ∈ W θ, vΦ+

i (λ) = Φ+i (λ) implies v ∈ W q, the Weyl group of the roots

perpendicular to ρi. Then by Proposition 4.16(a) v ∈ W q ⊂ W (G,H).Therefore w = uv−1 ∈ W (G,H). But w ∈ Wi so w ∈ W (G,H) ∩ Wi =W (M,H).

Since we are summing over cosets of W (M,H) we may as well assume

w = 1. Therefore γ1 = (H, Γ1,12(λ − µ)), γ2 = (H, Γ2,

12(λ − µ)). Since

these are both contained in LifteGG(γ), Γ1 and Γ2 agree on H0. Since they are

conjugate by g ∈ G they agree on Z(G). Since H = Z(G)H0 (Proposition4.7) this proves they are equal, proving the Lemma. �

Corollary 19.8 In the setting of the Theorem, let {γ1, . . . , γn} be the set of

constituents of LifteGG(wγ) as w runs over Wi, considered without multiplicity.

Then

(19.9) LifteGG(πst

G(γ)) = CG(H)

n∑

i=1

π eG(γi).

Remark 19.10 It is a remarkable fact that not only are the constituentsof (19.4) distinct, they have distinct central characters. See Remarks 11.16

94

and 12.9. As in [5] one can use this, together with Fourier inversion on Z(G)

to obtain a character formula for ΘeG(γ(w, i)). Since this uses a number of

structural results not needed elsewhere we omit the proof.

Corollary 19.11 Suppose π is an admissible virtual representation of G and

Θπ is stable. Then LifteGG(Θπ) is the character of a genuine virtual represen-

tation of G, or 0. If π is tempered and µ(χ, χs) ∈ iz then so is LifteGG(Θπ).

Every genuine standard module for G occurs in some Lift.

Lemma 19.12 Fix γ ∈ CDg(G). Define γ ∈ CD(G) as in Lemma 17.10, so

that γ occurs in LifteGG(γ). Then Θ eG(γ) occurs in Lift

eGG(Θst

G(γ)). Conversely,

suppose Θ eG(γ) occurs in LifteGG(Θst

G(γ′)) for some γ′ ∈ CD(G). Then Θst

G(γ) =

Θst

G(γ′).

Proof. The first statement is an immediate consequence of Theorem 19.1.

Now suppose that γ′ ∈ CD(G) such that Θ eG(γ) occurs in LifteGG(Θst

G(γ′)).

Then Θ eG(γ) = Θ eG(γ′) where γ′ occurs in LifteGG(wγ′) for some w ∈ W ′

i ,the imaginary Weyl group for the Cartan subgroup associated to γ′. SinceΘ eG(γ) = Θ eG(γ′), by Lemma 16.15 there is g ∈ G such that γ = gγ′. De-

fine g = p(p(g)). Then by Lemma 17.9 γ = gγ′ occurs in gLifteGG(wγ′) =

LifteGG(gwγ′). But by Lemma 17.10, this implies that γ = gwγ′. Hence

ΘstG(γ) = Θst

G(γ′). �

We now make the sum (19.4) more explicit. Assume LifteGG(Θst

G(γ)) 6= 0.

Then LifteGG(wγ) 6= ∅ for some w ∈ Wi. and after replacing γ with wγ we

assume w = 1. We may then describe the set of w ∈Wi such that LifteGG(wγ)

is non-empty. See (12.5) for a special case.

Fix γ = (H,Γ, λ) ∈ CD(G) and suppose LifteGG(πst

G(γ)) 6= 0. Without loss

of generality we may assume LifteGG(γ) 6= 0. Then each component of Lift

eGG(γ)

is of the form (H, Γ, 12(λ−µ)). In particular 1

2(λ−µ)−ρi(λ) is the differential

of a genuine character of H.Let

(19.13)(a) L = {X ∈ h | exp(X) ∈ Γ(H) ∩ C ∩H0}.

95

Note that Γ(H)∩C ∩H0 is a finite central subgroup, and L is a lattice. Let

(19.13)(b) W# = {w ∈Wi | exp((wλ/2 − λ/2)(X)) = 1 for all X ∈ L}.

Here is an alternative description of W# in terms of M = CentG(A). LetP ∨M be the weight lattice of the derived group of M , and

(19.13)(c) L′ = {γ∨ ∈ P ∨M | exp(2πiγ∨) ∈ Γ(H) ∩ C ∩H0}.

Note that XM∗⊂ L′ ⊂ P ∨

M where XM∗= X∗(T (C) ∩Md(C)). Then

(19.13)(d) W# = {w ∈Wi | 〈wλ/2− λ/2, γ∨〉 ∈ Z for all γ∨ ∈ L′}.

Suppose α ∈ Φi. By [3, Lemma 6.11]) and the fact that 12(λ− µ)− ρi(λ)

is the differential of a genuine character of H, we conclude 〈λ/2, α∨〉 ∈ Z orZ + 1

2depending on whether α is compact or noncompact. It follows that

W# ⊂ NormWi(Φi,c), and it is easy to see W (M,H) ⊂W#, so we have:

(19.14) W (Φi,c) ⊂W (M,H) ⊂W# ⊂ NormWi(Φi,c)

Compare [27, Proposition 4.16(d)].Frequently Γ(H) ∩ C ∩ H0 = 1, so L is the kernel of exp restricted to h

and L′ = XM∗.

Proposition 19.15 Fix γ ∈ CD(G) and suppose LifteGG(γ) 6= 0. With W#

as in (19.13) we have

(19.16) LifteGG(Θst

G(γ)) = CG(H)

∑

w∈W (M,H)\W#

∑

eγ∈LifteG

G(wγ)

Θ eG(γ).

There are |W (M,H)\W#||Z0(H)/φ(H)| terms in the sum.

For example suppose γ is a discrete series parameter for G = SL(2,R),and G = SL(2,R) or PSL(2,R). Then φ(H) = Z0(H) = H , W (G,H) = 1and W# = Z/2Z. See Example 18.26.

Proof. Fix w ∈ Wi. It is enough to show LifteGG(wγ) 6= 0 if and only if

w ∈W#, and by Lemma 17.11 this is equivalent to LiftfMM(wγ) 6= 0.

By assumption LifteHH(χ, χ,Γ) 6= ∅ where (χ, χ) ∈ S(H,Φ+, χs, χs). By

Proposition 10.11 Γ(h) = χ(h) for all h ∈ Ker(φH) where φH is the re-

striction of φ to H . Then wγ = (H,wΓ, wλ), and (cf. 18.3) LifteHH(wΓ) =

96

Lift(χ, χw, wΓ) where χw(h) = ewρ−ρ(h)χ(h). Thus LifteGG(wγ) 6= 0 if and

only if wΓ(h) = ewρ−ρ(h)χ(h) for all h ∈ Ker(φH). But wΓ(h) = Γ(w−1h) =χ(w−1h), so the condition is

(19.17)(a) χ((w−1h)h−1) = ewρ−ρ(h) for all h ∈ Ker(φH).

Write H = Z(M)(H ∩Md)0, and h = zh0 accordingly. Since z ∈ Z(M) and

w ∈ Wi = W (M(C), H(C)), (w−1h)h−1 ∈ (H ∩Md)0. By (6.2)(a) we can

write condition (a) as

(19.17)(b) (χ2eρ)((w−1h)h−1) = ewρ−ρ(h) for all h ∈ Ker(φH).

Let H(C)2 = {h ∈ H(C) | h2 = 1}. Then Ker(φH) = p(H(C)2) ∩ H , so wecan we can replace Ker(φH) with H(C)2 ∩ p−1(H) on the right hand side of(b), which then becomes equivalent to

(19.17)(c) χ2((w−1h)h−1) = 1 for all h ∈ H(C)2 ∩ p−1(H).

We claim this is equivalent to

(19.17)(d) χ2((w−1h)h−1) = 1 for all h ∈ H0, h2 ∈ Γ(H) ∩ C.

If h is in the set in (c), write h = ta with t ∈ exp(t(C)) and a ∈ exp(a(C)) =A(C). Then t is in the set in (d), and t = ha−1 with a ∈ A(C). Converselyfor h as in (d), suppose h2 = exp(iX) ∈ Γ(H) with X ∈ a (cf. 2.2) and leta = exp(iX/2) ∈ A(C). Then x = ha satisfies the condition in (c). Theclaim follows since Wi acts trivially on A(C).

For h in the set in (d) write h = exp(X/2) for X ∈ h. As in the discussion

preceding the Proposition choose χ to be a genuine character of Z(H) withdifferential 1

2(λ− µ) − ρi. Then since µ ∈ z(C)∗

(19.18)

χ2((w−1h)h−1) = exp((λ− µ− 2ρi)(w−1X/2 −X/2))

= exp(wλ/2 − λ/2)(X)exp((2ρi − w2ρi)(X/2))

= exp(wλ/2 − λ/2)(X)

The last equality follows from exp((2ρi−wρi)(X/2)) = exp((ρi−wρi)(X)) =1 since exp(X) = h2 ∈ Γ(H) ∩ C ⊂ Z(G). This gives (19.13)(b). Thealternative description ofW# is fairly standard. The kernel of the exponentialmap restricted to h is contained in t, and now everything is taking place inM , and in fact in Md. �

97

At least if φ(H) = Z0(H) the right hand side of (19.16), which is a sum

over W (M,H)\W♯, is a reasonable candidate for an L-packet for G. Unlikethe linear case the terms in this sum have different central characters; thisalso happens for Mp(2n,R) [4]. However W# may depend on G, so this sumis not canonical. We also note that there is no obvious notion of stability forgenuine virtual characters of G. Therefore the precise definition of L-packetfor G remains to be determined.

20 Appendix

In this section we extend Hirai’s Theorem on invariant eigendistributions toa class of groups containing all reductive groups of Harish-Chandra’s class.We first give some definitions from [17]. Let G be a reductive Lie groupwith real Lie algebra g. Let gd(C) be the derived algebra of the complex Liealgebra g(C). Let G∗(C) be the connected complex adjoint group of gd(C).

Definition 20.1 G satisfies condition A if the image of G under the adjointmap Ad is contained in G∗(C).

Recall [26] G is of Harish-Chandra’s class if G has finitely many connectedcomponents, Z(Gd) is finite, and condition A holds.

The kernel of Ad is CentG(G0), so let Ad(G) = G/CentG(G0). SinceZ(G) ⊂ CentG(G0) there is a natural map p : G/Z(G) → Ad(G).

Definition 20.2 G satisfies condition B if it satisfies condition A and thereis connected, complex group G1(C) with adjoint group G∗(C) and an injectivehomomorphism φ making the following diagram commute:

G/Z(G)φ //

p

��

G1(C)

Ad

��Ad(G)

Ad // G∗(C)

For example G satisfies condition B if it satisfies condition A and Z(G) =CentG(G0) (take G1(C) = G∗(C)). Note that GL(2,R) satisfies conditionB, but any admissible two-fold cover of GL(2,R) (cf. Section 3) satisfiescondition A (and is of Harish-Chandra’s class) but not condition B. The

98

failure of nonlinear groups to satisfy condition B requires us to extend Hirai’sresults.

Recall a connected complex Lie group is acceptable if ρ (one-half the sumof the positive roots) exponentiates to a character of a Cartan subgroup(cf. Section 2).

Definition 20.3 G is acceptable if it satisfies condition A and there is anacceptable, connected, complex group G1(C) with adjoint group G∗(C) and ahomomorphism φ making the following diagram commute:

Gφ //

p

��

G1(C)

Ad

��Ad(G)

Ad // G∗(C)

This definition of acceptability for disconnected groups is very strong. Forexample, GL(2,R) is not an acceptable group (there is no nontrivial homo-morphism from GL(2,R) to SL(2,C)), even though its identity componentis acceptable.

Hirai’s Theorem gives necessary and sufficient conditions for a class func-tion on G′ to be an invariant eigendistribution when G satisfies condition Band is acceptable. That the conditions are necessary requires only conditionA and acceptability, and is a fairly straightforward extension of results ofHarish-Chandra [10]. Hirai’s main contribution is to show that the condi-tions are sufficient, and for this he requires condition B.

Let G be a reductive Lie group satisfying condition A. Then the algebraof all left and right invariant differential operators on G is equal to thecenter Z of the universal envelopping algebra. In [16] Hirai showed how toextend the results of Harish-Chandra to show that when Θ is a G0-invarianteigendistribution for the action of Z, then Θ is given by integration against alocally integrable function on G which we also call Θ. Further, Θ is analyticon G′, the set of regular semisimple elements, and is a G0-invariant function,that is Θ(xgx−1) = Θ(g), x ∈ G0, g ∈ G′.

We assume for the remainder of this appendix that G is a reductive Liegroup satisfying condition A and use the notation of Section 13.

Definition 20.4 Let ν be a character of Z and let Θ be a G0-invariant func-tion on G′. We say Θ ∈ IE(ν) if Θ is the analytic function on G′ cor-

99

responding to a G0-invariant eigendistribution with infinitesimal characterν.

Let Θ be a G0-invariant function on G′. Let H be a Cartan subgroup ofG, Φ+ a choice of positive roots, and use the notation of Section 9. Define(20.5)

Ψ0(H,Φ+, h) = ∆0(Φ+, h)Θ(h), Ψ(H,Φ+, h) = ǫr(Φ+, h)Ψ0(H,Φ+, h).

Suppose that G is acceptable with φ : G → G1(C) satisfying the condi-tions of Definition 20.3. Let ρ = ρ(Φ+) and write ξρ(h) = eρ(φ(h)), h ∈ H .For h ∈ H ′ define(20.6)

Ψ0ρ(H,Φ

+, h) = ξρ(h)Ψ0(H,Φ+, h), Ψρ(H,Φ

+, h) = ξρ(h)Ψ(H,Φ+, h).

The functions Ψ0ρ(H,Φ

+) and Ψρ(H,Φ+) are the ones used by Hirai to state

his conditions. They can only be defined for acceptable groups. We firstrestate Hirai’s conditions in terms of the functions Ψ0(H,Φ+) and Ψ(H,Φ+)which can be defined without the assumption of acceptability.

For X ∈ h, define DX and DρX as in (13.1) and (13.2). The following

lemma follows easily from the definitions.

Lemma 20.7 Assume that G is acceptable. Let F ∈ C∞(H ′) and defineFρ(h) = ξρ(h)F (h), h ∈ H ′. Then

DFρ(h) = ξρ(h)DρF (h), h ∈ H ′, D ∈ S(h(C)).

Let F : H ′ → C, and let ν be a character of Z. Define conditions (C1,Φ+, ν) and (C2) as in Section 13. In the case that G is acceptable, we say Fsatisfies condition (CA1, ν) if it satisfies (C1, Φ+, ν) with Dρ replaced by D.

Lemma 20.8 1. If Ψ(H,Φ+) satisfies condition (C1, Φ+, ν) for one choiceof positive roots, then Ψ(H,Φ+) satisfies condition (C1, Φ+, ν) for anychoice of positive roots.

2. Assume that G is acceptable. Then Ψρ(H,Φ+) satisfies condition (CA1,

ν) if and only if Ψ(H,Φ+) satisfies condition (C1, Φ+, ν).

Proof. (1) Let Φ+ be any choice of positive roots. Since ǫr(Φ+, h) is lo-

cally constant on H ′, Ψ(H,Φ+) satisfies condition (C1, Φ+, ν) if and only ifΨ0(H,Φ+) satisfies condition (C1, Φ+, ν). Let w ∈W (Φ) so that wΦ+ is an-other choice of positive roots for Φ with ρ(wΦ+) = wρ. Then ∆0(wΦ+, h) =

100

ǫ(w)eρ−wρ(h)∆0(Φ+, h), h ∈ H , where ǫ(w) = ±1 is the determinant of w.Thus

(20.9) Ψ0(H,wΦ+, h) = ǫ(w)eρ−wρ(h)Ψ0(H,Φ+, h), h ∈ H ′.

Thus for any X ∈ h, h ∈ H ′,

DwρX Ψ0(H,wΦ+, h) = ǫ(w)eρ−wρ(h)Dρ

XΨ0(H,Φ+, h).

Thus Ψ0(H,wΦ+) satisfies condition (C1, wΦ+, ν) if and only if Ψ0(H,Φ+)satisfies condition (C1, Φ+, ν).

(2) follows from Lemma 20.7 and the fact that ξρ(h) and ξ−1ρ (h) are both

real analytic on H . �

The proof of the following lemma is similar to that of Lemma 20.8 sinceǫr(Φ

+, h) is constant on each connected component of H ′(R).

Lemma 20.10 1. If Ψ(H,Φ+) satisfies condition (C2) for one choice ofpositive roots, then Ψ(H,Φ+) satisfies condition (C2) for every choiceof positive roots.

2. Assume that G is acceptable. Then Ψρ(H,Φ+) satisfies condition (C2)

if and only if Ψ(H,Φ+) satisfies condition (C2).

Let α ∈ Φ+r and define J,Φ+

J = cΦ+, H(α), and β = c∗(α) as in Section13. Let h ∈ H(α) and assume that Ψ(H,Φ+) satisfies condition (C1, Φ+, ν)and (C2). Then Dρ,±

α Ψ(H,Φ+, h) are defined as in (13.4). There are alsowell-defined limits

(20.11) DραΨ

0(H,Φ+, h) = lims→0

DραΨ

0(H,Φ+, hexp(sα)).

and when G is acceptable

(20.12) DαΨ0ρ(H,Φ

+, h) = lims→0

DαΨ0ρ(H, hexp(sα)).

Further, h ∈ J ′(R), so that DρJ

βΨ0(J,Φ+

J , h) and DρJ

βΨ(J,Φ+

J , h) are defined,

and when G is acceptable, DβΨ0ρJ

(J,Φ+J , h) is defined.

Define

(20.13) ǫαr (Φ+, h) = sign

∏

γ∈Φ+r ,γ 6=α

(1 − e−γ(h)), h ∈ H(α).

101

Lemma 20.14 For all h ∈ H(α),

Dρ,±α Ψ(H,Φ+, h) = ±ǫαr (Φ+, h)Dρ

αΨ0(H,Φ+, h);

DρJ

βΨ(J,Φ+

J , h) = ǫr(Φ+J , h)D

ρJ

βΨ0(J,Φ+

J , h).

Proof. Fix h ∈ H(α). The first equation holds because for all small s 6= 0,

ǫr(Φ+, hexp(sα)) = ǫαr (Φ

+, h)sign(s).

The second equation holds because h ∈ J ′(R) so that ǫr(Φ+J , h) is constant

in a neighborhood of h. �

Lemma 20.15 Assume that Φ+ is a choice of positive roots such that α isa simple root for Φ+

r . Then ǫαr (Φ+, h) = ǫr(Φ

+J , h) for all h ∈ H(α).

Proof. Fix h ∈ H(α). Write Φ+r \{α} = Φ1 ∪ Φ2 where

Φ1 = {γ ∈ Φ+r :< γ, α >= 0}

and Φ2 is its complement in Φ+r \{α}. Then the positive real roots in ΦJ are

given by Φ+r,J = c∗Φ1. Thus

ǫr(Φ+J , h) = sign

∏

γ∈Φ1

(1 − e−c∗γ(h)) = sign

∏

γ∈Φ1

(1 − e−γ(h))

since ec∗γ(h) = eγ(h) for all γ ∈ Φ, h ∈ H(α). Thus

ǫαr (Φ+, h) = ǫr(Φ

+J , h)sign

∏

γ∈Φ2

(1 − e−γ(h)).

Let γ ∈ Φ2. Then since α is simple for Φ+r , γ 6= α, and < γ, α > 6= 0,

sαγ ∈ Φ2, sαγ 6= γ. But e−sαγ(h) = e−γ(h). Thus

sign∏

γ∈Φ2

(1 − e−γ(h)) = 1.

�

Consider the following conditions.

(20.16) DαΨ0ρ(H,Φ

+, h) = DβΨ0ρJ

(J,Φ+J , h), h ∈ H(α).

102

(20.17) DραΨ

0(H,Φ+, h) = DρJ

βΨ0(J,Φ+

J , h), h ∈ H(α).

Condition (20.16) is the condition used by Hirai in [16] for acceptable groups,and (20.17) is the analogous condition for not necessarily acceptable groups.It is useful for our applications in Section 14 to replace (20.17) by condition(C3) of Section 13. The following lemma clarifies the relationship betweenthese conditions.

Lemma 20.18 1. If Ψ0(H,Φ+) satisfies (20.17) for one choice of positiveroots, then Ψ0(H,Φ+) satisfies (20.17) for every choice of positive roots.

2. Let Φ+ be a choice of positive roots such that α is simple for Φ+r . Then

Ψ0(H,Φ+) satisfies (20.17) if and only if Ψ(H,Φ+) satisfies condition(C3).

3. Assume that G is acceptable. Then Ψ0ρ(H,Φ

+) satisfies (20.16) if andonly if Ψ0(H,Φ+) satisfies (20.17).

Proof. Let h ∈ H(α) and let Φ+ be any choice of positive roots.(1) Let w ∈ W (Φ). Then wJ = c∗w(c∗)−1 ∈W (ΦJ), c

∗wΦ+ = wJΦ+J and

wJρJ = ρ(wJΦ+J ). Further,

Dwρα Ψ0(H,wΦ+, h) = ǫ(w)eρ−wρ(h)Dρ

αΨ0(H,Φ+, h);

DwJρJ

βΨ0(J, wJΦ

+J , h) = ǫ(wJ)e

ρJ−wJρJ (h)DρJ

βΨ0(J,Φ+

J , h).

But ǫ(w) = ǫ(wJ) and eρJ−wJρJ (h) = ec∗(ρ−wρ)(h) = eρ−wρ(h). Thus Ψ0(H,wΦ+)

satisfies (20.17) when Ψ0(H,Φ+) does.(2) follows from combining Lemmas 20.14 and 20.15.(3) Using Lemma 20.7 we have

DαΨ0ρ(H,Φ

+, h) = ξρ(h)DραΨ

0(H,Φ+, h);

DβΨ0ρJ

(J,Φ+J , h) = ξρJ

(h)DρJ

βΨ0(J,Φ+

J , h).

Recall that ξρ, ξρJare defined using a homomorphism φ : G → G1(C). Let

H1(C) and J1(C) be the Cartan subgroups of G1(C) with Lie algebras h(C)and j(C). Then we can pick a representative of the Cayley transform c ∈G1(C) with cH1(C)c−1 = J1(C) and chc−1 = h for all h ∈ H1(C) ∩ J1(C).Then for h ∈ H(α),

ξρJ(h) = eρJ (φ(h)) = ec

∗ρ(φ(h)) = eρ(cφ(h)c−1) = eρ(φ(h)) = ξρ(h),

103

since φ(h) ∈ H1(C) ∩ J1(C). Thus Ψ0ρ(H,Φ

+) satisfies (20.16) if and only ifΨ0(H,Φ+) satisfies (20.17). �

Definition 20.19 Θ satisfies condition C(ν) if for every Cartan subgroup Hof G and choice of positive roots Φ+, Ψ(H,Φ+) satisfies conditions (C1,Φ+, ν)and (C2), and Ψ0(H,Φ+) satisfies (20.17) for all α ∈ Φ+

r .

Hirai’s necessary and sufficient conditions are (CA1, ν), (C2), and (20.16).Using Lemmas 20.8, 20.10, and 20.18, when G is acceptable, Θ satisfies C(ν)if and only if Θ satisfies Hirai’s conditions. Thus we can state Hirai’s Theoremas follows.

Theorem 20.20 (Hirai, [17]) Let G be a reductive Lie group that satisfiescondition B and is acceptable. Let Θ be a G0-invariant function on G′ andlet ν be a character of Z. Then Θ ∈ IE(ν) if and only if Θ satisfies C(ν).

Remark 20.21 Suppose that Θ is a G-invariant function on G′. If Θ ∈IE(ν), then the corresponding eigendistribution is G-invariant. Thus Θ cor-responds to a G-invariant eigendistribution if and only if Θ satisfies C(ν).

Remark 20.22 In Hirai’s statement of Theorem 20.20 in §11 of [17], hestarts with the functions Ψρ(H,Φ

+) (which he calls κj) on each of a set ofrepresentatives Hj of G0-conjugacy classes of Cartan subgroups. He thengives an extra condition that he calls ǫ-symmetric which guarantees thatthey can be patched together to give a G0-invariant function on G′ which isour Θ. We don’t use this condition since we assume from the beginning thatwe have a G0-invariant function on G′.

We will prove the following extension of Hirai’s Theorem.

Theorem 20.23 Let G be a reductive Lie group which satisfies condition A.Let Θ be a G0-invariant function on G′ and let ν be a character of Z. ThenΘ ∈ IE(ν) if and only if Θ satisfies C(ν).

We will prove Theorem 20.23 by making a number of reductions and thenapplying Hirai’s theorem. A number of routine lemmas are stated withoutproof.

Let G be a reductive Lie group which satisfies condition A. Let Θ be a G0-invariant function on G′ and let ν be a character of Z. Since G is a Lie group,G has at most countably many connected components. Write G = ∪ixiG0,

104

and let Θi = Θχi where χi is the characteristic function of xiG0. Then each

Θi is a G0-invariant function on G′ which is supported on xiG0.

Lemma 20.24 Θ ∈ IE(ν) if and only if Θi ∈ IE(ν) for all i. Further, Θsatisfies C(ν) if and only if Θi satisfies C(ν) for all i.

Lemma 20.25 Each connected component of G has a representative x suchthat x2 ∈ CG(G0) and < x > ∩CG(G0)G0 ⊂ CG(G0) where < x > denotesthe cyclic subgroup generated by x.

Proof. Let G∗(C) denote the connected complex adjoint group with Liealgebra gd(C). Since G satisfies condition A there is a homomorphism p :G → G∗(C) with kernel CG(G0) such that Ad p(g) = Ad g, g ∈ G. In fact,p(G) ⊂ G∗(R), the real points of G∗(C). Fix x ∈ G. Then p(xG0) is aconnected component of G∗(R). If p(xG0) = (G∗)0, then xG0 ⊂ CG(G0)G0

so we can pick our representative x ∈ CG(G0). Otherwise, p(xG0) = t(G∗)0

where t2 = 1 and t 6∈ (G∗)0. We can pick our representative x so thatp(x) = t. Then x 6∈ CG(G0)G0, but p(x2) = 1 so x2 ∈ CG(G0). Thus< x > ∩CG(G0)G0 =< x2 >⊂ CG(G0). �

First Reduction. Because of Lemmas 20.24 and 20.25 we may as wellassume that Θ is supported on xG0 where x ∈ G such that x2 ∈ CG(G0) and< x > ∩CG(G0)G0 ⊂ CG(G0).

Fix x ∈ G such that x2 ∈ CG(G0) and < x > ∩CG(G0)G0 ⊂ CG(G0).Define τ : G0 → G0 by τ(g) = xgx−1, g ∈ G0. Since x2 ∈ CG(G0) we haveτ 2 = 1. Let Gτ =< τ > ⋉G0 be the semidirect product of < τ >= {1, τ}and G0. That is, if x ∈ CG(G0), then τ = 1 and Gτ = G0. OtherwiseGτ = G0 ∪ τG0 has two connected components.

Define φ : xG0 → τG0 by φ(xg) = τg, g ∈ G0. Then φ is a diffeomorphismwith Ad(xg) = Ad(φ(xg)) for all g ∈ G0 and φ(hxgh−1) = hφ(xg)h−1 for allh, g ∈ G0. Thus there is a bijection between G0-invariant functions Θ on G′

which are supported on xG0 and G0-invariant functions Θτ on G′τ which are

supported on τG0 given by Θ(xg) = Θτ (φ(xg)), g ∈ G0.

Lemma 20.26 Let Θ be a G0-invariant function on G′ which is supportedon xG0, and let Θτ be the corresponding G0-invariant function on G′

τ whichis supported on τG0. Then Θ ∈ IE(ν) as a function on G′ if and only ifΘτ ∈ IE(ν) as a function on G′

τ . Further, Θ satisfies C(ν) if and only if Θτ

satisfies C(ν).

105

Lemma 20.27 CGτ(G0) = Z(G0).

Proof. Clearly CGτ(G0) ∩ G0 = Z(G0). Suppose there is g ∈ G0 such that

τg ∈ CGτ(G0). Then τ ∈ CGτ

(G0)G0 so that x ∈< x > ∩CG(G0)G0 ⊂CG(G0) and τ = 1. Thus τG0 ∩ CGτ

(G0) = ∅ unless τG0 = G0. In any casewe have CGτ

(G0) = CGτ(G0) ∩G0 = Z(G0). �

Second Reduction. Because of Lemma 20.26 we may as well replace Gby Gτ . Thus we assume there is x ∈ G with x2 = 1 and G =< x > ⋉G0.Moreover, because of Lemma 20.27 we have CG(G0) = Z(G0). We continueto write τ(g) = xgx−1, g ∈ G0.

Let C be a closed discrete subgroup of Z(G0) such that τ(C) = C. ThenC is a normal subgroup of G. Define G1 = G/C and x1 = xC. ThenG0

1 = G0/C, x21 = 1, and G1 =< x1 > ⋉G0

1.Suppose that Θ1 is a G0

1-invariant function on G1. Then Θ1 lifts to aG0-invariant function Θ on G′

Lemma 20.28 Θ ∈ IE(ν) if and only if Θ1 ∈ IE(ν). Further, Θ satisfiesC(ν) if and only if Θ1 satisfies C(ν).

Lemma 20.29 Let C = {τ(z)z−1 : z ∈ Z(G0)}. Then C is a closed discretesubgroup of Z(G0) with τ(C) = C. Define G1 = G/C. Then Z(G1) =CG1

(G01) = Z(G0

1).

Proof. Define ψ : Z(G0) → Z(G0) by ψ(z) = τ(z)z−1. Then ψ is ahomomorphism so that C = φ(Z(G0)) is a subgroup of Z(G0). Writeg = gd ⊕ z where gd = [g, g] and z is the center of g. Then Z(G0) = Z(G0

d)Zwhere Z = expz. Now Z(G0

d) is a closed discrete subgroup of Z(G0) andτ(z) = z for all z ∈ Z. Thus C = {τ(z)z−1 : z ∈ Z(G0

d)} ⊂ Z(G0d) and

so is a closed discrete subgroup of Z(G0). Further, since τ 2 = 1, we haveτ(ψ(z)) = ψ(z−1) ∈ C for all z ∈ Z(G0). Thus τ(C) = C.

Write p : G → G1 = G/C for the projection. Then Z(G01) = p(Z(G0)).

Let z ∈ Z(G0). Then τ(z)z−1 ∈ C so that p(x)p(z)p(x)−1 = p(τ(z)) = p(z).Thus p(z) ∈ Z(G1). Thus Z(G0

1) ⊂ Z(G1) ⊂ CG1(G0

1). Conversely, letg ∈ G and suppose that p(g) ∈ CG1

(G01). Then Ad g = Ad p(g) = 1 so

that g ∈ CG(G0) = Z(G0) by Lemma 20.27. Thus p(g) ∈ Z(G01). Thus

CG1(G0

1) ⊂ Z(G01). �

106

Lemma 20.30 Define C as in Lemma 20.29. Let Θ be a G0-invariant func-tion on G′ which is supported on xG0. Then

Θ(gc) = Θ(g), g ∈ G′, c ∈ C.

That is, Θ factors to a G01-invariant function Θ1 on G′

1 supported on x1G01.

Proof. Since Θ is supported on xG0 and C ⊂ G0, the result is true ifg 6∈ xG0. Let g ∈ G0 with xg ∈ G′ and let c ∈ C. Then there is z ∈ Z(G0)such that c = τ(z)z−1 = xzx−1z−1 = x−1zxz−1. Since Θ is G0-invariant andx−1zx ∈ Z(G0) we have

Θ(xg) = Θ(zxgz−1) = Θ(xx−1zxgz−1) = Θ(xgx−1zxz−1) = Θ(xgc).

�

Third Reduction. By Lemmas 20.28 and 20.30 we may as well assumethat G = G1. But by Lemma 20.29 we have Z(G1) = CG1

(G01) = Z(G0

1).That is, we can assume that Z(G) = CG(G0) = Z(G0) and there is x ∈ Gwith x2 = 1 and G =< x > ⋉G0.

The following Lemma completes the proof of Theorem 20.23.

Lemma 20.31 Let G be a reductive Lie group satisfying condition A. As-sume that Z(G) = CG(G0) = Z(G0) and there is x ∈ G with x2 = 1 andG =< x > ⋉G0. Let Θ be a G0-invariant function on G′ and let ν be acharacter of Z. Then Θ ∈ IE(ν) if and only if Θ satisfies C(ν).

Proof. Write g = gd⊕ z as in Lemma 20.29. Then G0 = G0dZ where G0

d andZ are the connected subgroups of G corresponding to gd and z respectively.Since G0

d is a connected semisimple Lie group there is a connected finitecentral extension p1 : G1 → G0

d such that G1 is acceptable. That is, there isa connected acceptable complex Lie group G2(C) with Lie algebra gd(C) suchthat the inclusion map of gd ⊂ gd(C) lifts to a homomorphism φ1 : G1 →G2(C). Let K = ker p1 ∩ ker φ1 ⊂ Z(G1) and define G1 = G1/K. Then p1

and φ1 both factor toG1, so that we have p1 : G1 → G0d and φ1 : G1 → G2(C).

Thus G1 is also an acceptable cover of G0d. Thus we may as well assume that

ker p1 ∩ ker φ1 = {1}.Since G satisfies condition A and G2(C) is a complex Lie group with

Lie algebra gd(C) there is x2 ∈ G2(C) such that Ad x = Ad x2. ThenAd x2

2 = Ad x2 = 1 so that x22 ∈ Z(G2(C)). Thus x2 has finite order. Let

107

p2 : G1 → G1 be the simply connected cover of G1. Then there is a uniqueautomorphism τ of G1 such that

τ(expX) = exp(Ad xX) = exp(Ad x2X), X ∈ g.

It satisfies τ 2 = 1, and

p1p2τ(g) = xp1p2(g)x−1, φ1p2τ(g) = x2φ1p2(g)x

−12 , g ∈ G1.

Suppose that g ∈ ker p2. Then

p1p2τ(g) = xp1p2(g)x−1 = 1, φ1p2τ(g) = x2φ1p2(g)x

−12 = 1.

Thus p2τ(g) ∈ ker p1∩ker φ1 = {1}, so that τ(g) ∈ ker p2. Thus τ descendsto give a well-defined automorphism τ : G1 → G1 satisfying τ 2 = 1 and

(20.32) p1(τ(g1)) = xp1(g1)x−1, φ1(τ(g1)) = x2φ1(g1)x

−12 , g1 ∈ G1.

Assume that z1 ∈ Z(G1). Then p1(z1) ∈ Z(G0d) ⊂ Z(G0) = Z(G) by as-

sumption. Further, Ad(φ1(z1)) = Ad z1 = 1. This implies that φ1(z1) ∈Z(G2(C)) since G2(C) is connected. Thus p1(τ(z1)) = xp1(z1)x

−1 = p1(z1)and φ1(τ(z1)) = x2φ1(z1)x

−12 = φ1(z1). Thus τ(z1)z

−11 ∈ ker p1 ∩ ker φ1 =

{1}. That is, τ(z1) = z1 for all z1 ∈ Z(G1).Let m = 2m2 where m2 is the order of x2 and write Zm for the additive

group of integers mod m. Let G = {(z, k, g) : z ∈ Z, k ∈ Zm, g ∈ G1}. Forz1, z2 ∈ Z, k, q ∈ Zm, g1, g2 ∈ G1, define the product

(20.33) (z1, k, g1)(z2, q, g2) = (z1z2, k + q, τ−q(g1)g2).

Thus G is the direct product of Z with the semidirect product of Zm and G1

where k ∈ Zm acts on G1 by τk. G0 is isomorphic to the direct product of Zand G1 and so G is a reductive Lie group with Lie algebra g. We calculatethat

(20.34) (z, k, g)(z2, q, g2)(z, k, g)−1 = (z2, q, τ

k(τ−q(g)g2g−1)).

Thus Ad(z, k, g) = Ad(xkp1(g)) so that G satisfies condition A. We will showthat

(20.35) Z(G) = C eG(G0) = {(z, k, g) : z ∈ Z, xk = 1, g ∈ Z(G1)}.

108

Using (20.34) and the fact that τ(g) = g for all g ∈ Z(G1), it is clear that

{(z, k, g) : z ∈ Z, xk = 1, g ∈ Z(G1)} ⊂ Z(G) ⊂ C eG(G0). Suppose that

(z, k, g) ∈ C eG(G0). Then Ad(xkp1(g)) = Ad(z, k, g) = 1, so that xkp1(g) ∈CG(G0) = Z(G) = Z(G0) by assumption. Thus xk ∈< x > ∩G0 = {1}.Thus xk = 1 and p1(g) ∈ Z(G) ∩ G0

d = Z(G0d) so that g ∈ Z(G1). Thus

C eG(G0) ⊂ {(z, k, g) : z ∈ Z, xk = 1, g ∈ Z(G1)} ⊂ Z(G). This shows that Gsatisfies condition B.

Define φ : G → G2(C) by φ(z, k, g) = xk2φ1(g). It is well-defined becausem is a multiple of the order of x2. It is easy to show it is a homomorphismusing (20.32). It is a continuous homomorphism which induces canonically

the natural injection of Ad(G) into Ad (G2(C)). Thus G is acceptable.

Define p : G → G by p(z, k, g) = zxkp1(g). The mapping is well-definedsince m is even, and it satisfies Ad p(z, k, g) = Ad(z, k, g). It is easy toshow that it is homomorphism using (20.32), and it is clearly surjective. Let(z, k, g) ∈ ker p so that zxkp1(g) = 1. Then Ad(z, k, g) = Adp(z, k, g) = 1 so

that (z, k, g) ∈ C eG(G0) = Z(G). Thus ker p ⊂ Z(G). Further, using (20.35)

we have Z(G)0 = {(z, 0, 1) : z ∈ Z}. Thus the identity component of ker p is

contained in Z(G)0 ∩ ker p = {(1, 0, 1)} so that ker p is a discrete subgroup

of Z(G).Let Θ be a G0-invariant function on G′ and let ν be a character of Z.

Lift Θ to a G0-invariant function Θ on G′. Then Θ ∈ IE(ν) if and only

if Θ ∈ IE(ν) and Θ satisfies C(ν) if and only if Θ satisfies C(ν). But G

is acceptable and satisfies condition B. Thus Θ ∈ IE(ν) if and only if Θsatisfies C(ν) by Theorem 20.20. �

References

[1] J. Adams. Unitary shimura correspondence for split real groups. Journalof the American Mathematical Society, 20(3). preprint.

[2] J. Adams. Characters of non-linear groups. Proceedings of Conf. onRepresentation Theory and Harmonic Analysis, Okayama, Japan, 26:1–18, 2000.

[3] J. Adams and P. Trapa. Duality for nonlinear simply laced groups.preprint, http://www.math.umd.edu/∼jda/preprints/duality.pdf

109

[4] Jeffrey Adams. Lifting of characters on orthogonal and metaplecticgroups. Duke Math. J., 92(1):129–178, 1998.

[5] Jeffrey Adams. Characters of covering groups of SL(n). J. Inst. Math.Jussieu, 2(1):1–21, 2003.

[6] Jeffrey Adams. Nonlinear covers of real groups. Int. Math. Res. Not.,(75):4031–4047, 2004.

[7] Jeffrey Adams, Dan Barbasch, and David A. Vogan, Jr. The Langlandsclassification and irreducible characters for real reductive groups, volume104 of Progress in Mathematics. Birkhauser Boston Inc., Boston, MA,1992.

[8] Y. Flicker D. A. Kazhdan. Metaplectic correspondence. 64:53–110, 1986.

[9] Y. Flicker. Automorphic forms on covering groups of GL(2). 57:119–182,1980.

[10] Harish-Chandra. Invariant eigendistributions on a semisimple Lie group.Trans. Amer. Math. Soc., 119:457–508, 1965.

[11] Harish-Chandra. Harmonic analysis on real reductive groups. I. Thetheory of the constant term. J. Functional Analysis, 19:104–204, 1975.

[12] Harish-Chandra. Supertempered distributions on real reductive groups.In Studies in applied mathematics, volume 8 of Adv. Math. Suppl. Stud.,pages 139–153. Academic Press, New York, 1983.

[13] Henryk Hecht and Wilfried Schmid. Characters, asymptotics and n-homology of Harish-Chandra modules. Acta Math., 151(1-2):49–151,1983.

[14] Rebecca A. Herb. Two-structures and discrete series character formulas.In The mathematical legacy of Harish-Chandra (Baltimore, MD, 1998),pages 285–319. Amer. Math. Soc., Providence, RI, 2000.

[15] Takeshi Hirai. Invariant eigendistributions of Laplace operators on realsimple Lie groups. I. Case of SU(p, q).. Japan. J. Math., 39:1–68, 1970.

110

[16] Takeshi Hirai. Supplements and corrections to my paper: “The char-acters of some induced representations of semisimple Lie groups” (J.Math. Kyoto Univ. 8 (1968), 313–363). J. Math. Kyoto Univ., 15:237–250, 1975.

[17] Takeshi Hirai. Invariant eigendistributions of Laplace operators on realsimple Lie groups. II. General theory for semisimple Lie groups. Japan.J. Math. (N.S.), 2(1):27–89, 1976.

[18] J.-S. Huang. The unitary dual of the universal covering group ofGL(n,R). 61:705–745, 1990.

[19] J.-S. Huang J. Adams. Kazhdan-patterson lifting for GL(n,R).89(3):423–444, 1997.

[20] D. A. Kazhdan and S. J. Patterson. Metaplectic forms. Inst. HautesEtudes Sci. Publ. Math., (59):35–142, 1984.

[21] D. A. Kazhdan and S. J. Patterson. Towards a generalized Shimuracorrespondence. Adv. in Math., 60(2):161–234, 1986.

[22] A. Knapp. Representation Theory of Semisimple Groups. An overviewbased on Examples. Princeton University Press, Princeton, NJ, 1986.

[23] David Renard. Endoscopy for Mp(2n,R). Amer. J. Math., 121(6):1215–1243, 1999.

[24] D. Shelstad. Characters and inner forms of a quasi-split group over R.Compositio Math., 39(1):11–45, 1979.

[25] D. Shelstad. l-indistinguishability for real groups. 259:385–430, 1982.

[26] D. Vogan. Representations of Real Reductive Lie Groups, volume 15 ofProgress in mathematics. Birkhauser, Boston, 1981.

[27] D. Vogan. Irreducible characters of semisimple Lie groups IV. character-multiplicity duality. 49, No. 4:943–1073, 1982.

[28] David A. Vogan, Jr. Unitarizability of certain series of representations.Ann. of Math. (2), 120(1):141–187, 1984.

111